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10 Really Cool Jozi Places I Didn't Blog About in 2017
by 2summers | Jan 2, 2018 | Arts and Culture, Food and Drink, Johannesburg, Markets/Shopping, Townships/Informal Settlements | 4 comments
I blogged about a lot of Jozi places in 2017. But there are also a lot of places I visited but never managed to write up, for one reason or another.
Lots of these places are really cool and I don't want their photos to languish forever in my Lightroom catalog. Also, I want you to visit them. So here is a quick rundown of the 10 best places I never blogged about in 2017.
Jozi Places I Should Have Blogged About in 2017
1) Urban Backpackers/Kafe Noir (Joburg CBD)
Urban Backpackers is a youth hostel smack in the middle of the Joburg CBD, at 98 Anderson Street. The ground floor of Urban Backpackers is Kafe Noir, a funky coffeeshop. The rooftop is a bar and party venue.
Chilling on the rooftop of Urban Backpackers after the Hi-Tec #WalkMyCity event. (Photo: Fiver Löcker)
Urban Backpackers is owned by Papy Nakuw, a fashion designer who also owns the Urban Zulu clothing label. Papy is impossibly cool. His design workshop is just across the street.
Papy at Urban Backpackers.
I like the Urban Backpackers because it's an unexpected place in an unexpected location. Kafe Noir and the rooftop are generally open to the public but you should call first to be sure.
2) The Royale (Craighall)
The Royale is a Cuban-inspired restaurant, occupying the spot where Warm & Glad used to be. The Royale is notable for its colorful, Caribbean-inspired decor, which seems to be a trend in the Joburg restaurant scene at the moment.
The Royale bar.
Hip new restaurants popped up around Joburg at an alarming rate in 2017 — each one more trendy and glamorous than the next. I visited many and blogged about few. But the Royale was one of my favorites. The Cuban sandwich I ordered was tasty and the rum cocktails were strong. When I was there in late November, a speakeasy was in progress in a back room behind the restaurant. Hopefully it's opened by now.
Neon sign at the Royale entrance, which I think was designed specifically with Instagram in mind.
3) Manny's – Hillside Fish & Chips (Rosettenville)
Manny's, which is also called Hillside Fish & Chips (I'm not sure which name is paramount but I'm going with Manny's), is the opposite of the Royale. It's not new, nor is it trendy. Manny's has been around for decades and according to my friend Kate, who frequented Manny's as a kid, it hasn't changed one bit.
The staff at Manny's. I seem to have lost their names but the man isn't Manny.
I ordered a sandwich frighteningly named "the pregnant prego roll". I can't remember exactly what was on it but there was definitely meat (chicken, beef, or maybe both), a fried egg, slap chips (soggy french fries), and a lot of sauce. Maybe some cheese, too. It was delicious and cost R45 ($3.60).
The Pregnant Prego Roll. Perhaps so-named because eating it gives you a food baby.
4) Rio Douro Fisheries (Rosettenville)
Rio Douro Fisheries is a legendary fishmonger and butcher in the South of Joburg. I went to Rio Douro on the same day I went to Manny's, on a tour of the South with Kate, and intended to write them up together as a #Gauteng52 post. Somehow it never happened.
Rio Douro.
Rio Douro is worth visiting for its large selection of fish, which is a Joburg rarity. Rio Douro also sells lots of interesting Portuguese foods and cooking accessories.
Best of all, Rio Douro has the coolest security guards in town.
Patrick, a cool Rio Douro security guard, in front of an equally cool ocean-themed mural.
5) Linden Swimming Pool (Linden)
I also considered including the Linden Pool in #Gauteng52, but I'd already done the Zoo Lake Pool and one Joburg swimming pool seemed enough.
But the Linden pool is particularly cool, first because it's one of the few indoor public pools in the city and also for its interesting architecture.
The covered Linden Swimming Pool, whose roof looks vaguely spacecraft-like.
The Linden Pool is popular on rainy Saturdays.
I visited the Linden Pool as part of a group project called #20Laps, which you can read more about on my friend Ang's blog, Jozi.Rediscovered. I think Ang's Linden Pool post is coming up next.
6) Agog (New Doornfontein)
I think I held off on writing about Agog because it's difficult to describe. It's a wine bar, whiskey bar, restaurant, and art gallery, with multiple floors and a rooftop with interesting views of downtown Joburg. Agog is part of the Maboneng Precinct, in a formerly industrial section of New Doornfontein.
Marie-Lais and Patrick, who works at Agog, enjoy a bottle of wine.
Agog's ground floor is home to Nine Barrels, the wine bar, which features a huge variety of local and international wines by the glass. The middle floors house art exhibitions. The top floor is a whiskey and gin bar.
A view to the north from Agog, featuring Ellis Park stadium and an evening commuter train.
Good wine by the glass is hard to find in Joburg, and Agog is the best place to find it. It's also just a lovely spot to spend an afternoon in the city.
7) Tembisa
Tembisa is a large township in Ekurhuleni, Joburg's East Rand.
I included Winnie's Tuck Shop, a kota restaurant in Tembisa, in the #Gauteng52 series. But I never got around to writing about Tembisa as a whole, which I experienced with a local tour guide named Walter Msibi.
Tembisa has an interesting history. It used to be an Afrikaans family farm and there are some buildings still standing from the original farmstead.
A farmer growing cabbages and spinach in the middle of Tembisa, on a plot irrigated from a natural spring.
I can't go into Tembisa's full history here, but I encourage you to book a tour with Walter. His number is 078-248-8598. You can also book Walter through Mozee Tours and Transport.
Natural spring water flows through Tembisa.
8) The Stuff We Love (Melville)
The Stuff We Love used to be a vintage shop on 7th Street in Melville. But the owners, Susan and Gwynn, recently closed the shop and went mobile, selling their beautiful clothes at various pop-ups in Melville and surrounds. I visited one such pop-up a few months ago at Poppy's Restaurant and had an insanely fun time.
Susan (left) and Gwynn (right) at their Poppy's pop-up.
Susan and Gwynn are fabulous. The clothes they sell are fabulous. Their customers are all fabulous. And they always have champagne.
I didn't buy this skirt but maybe I should have. (Photo: Marie-Lais Emond)
The Stuff We Love is working on a website so they can sell online. In the meantime, follow their movements on Facebook.
9) Victoria Yards (Lorentzville)
Victoria Yards is also difficult to explain. It's a brand-new complex in Lorentzville, which is between Bertrams and Troyeville, developed from a sprawl of old industrial buildings. Victoria Yards currently houses artists' galleries and studios, event space, an urban garden, a chair factory, and a brewery/pizza parlor. There's new stuff happening all the time.
Victoria Yards.
Crazy chairs at an art gallery in Victoria Yards.
I've got a couple of smaller posts planned about various businesses in Victoria Yards, but for now I think the whole place deserves a mention.
10) The Jazzfarm (Lanseria)
I hesitated to include the Jazzfarm as a #Gauteng52 entry because it is not a public place, per se. But it's too interesting not to mention here.
The Jazzfarm is home to Myrtle Clarke and Jules Stobbs, aka the Dagga Couple. I found my way to the farm with Marie-Lais to do a story about the Dagga Couple's labyrinth, which is extremely beautiful.
A labyrinth is kind of like a maze, except you can't get lost because there is always just one direction to walk in. I loved it.
The Dagga Couple don't publicize the location of their farm for obvious reasons. But they do have a Facebook page and they said people are welcome to come by appointment to walk the labyrinth. I recommend this. Also, it's a great activity to combine with lunch at the nearby Culinary Table.
Thanks again to the friends who accompanied me to these places and especially to Marie-Lais, whose "Other Side of the City" column in the Saturday Citizen newspaper — Marie-Lais writes the column and I take the photos — motivated several of the visits.
There will be many more stories in 2018.
Floating around at the Jazzfarm. (Photo: Marie-Lais Emond)
catji	on January 3, 2018 at 5:42 am
Wow, all that and The Dagga Couple too. 🙂
2summers	on January 3, 2018 at 2:47 pm
I know, right? It was that kind of year 🙂
Kevin k m Collins	on January 3, 2018 at 2:16 pm
Thanks for the update – always love your blog. Please do come back to the REEA farm in Craighall as the Gerakaris Winery is open, licenced and running tastings – new grapes arrive in Feb for a winemaking and would be lovely to show you around again.
What? I had no idea there was a winery in Craghall! I'd love to come.


Ahmedabad is ranked third in the list of the fastest growing cities of the decade by the Forbes magazine. It is the fifth largest city in incredible India and has a population of about 6.7 million. The pace at which the city has developed is phenomenal. A few years back it was a small city with very less infrastructure and not much development. Today, it is expanding and growing like a fire. The super-efficient and fast BRTS bus system has made transportation a breezy smooth process. One should never hesitate in investing in new flats in Ahmedabad as the city is vast and there is great scope of development and growth!
The recent years saw the emergence of the city as an important economic and commercial hub. It ranks second when it comes to producing cotton in India. The gross domestic product of India was a whopping 64 billion in 2014! It is just 30 kilometers away from the state capital of Gujarat, Gandhinagar. These facts prove that investing in Ahmedabad's real estate would be beneficial and new projects in Ahmedabad would never fail or fall.
Ahmedabad is an important tourist attraction as it has many attractions such as the Sabarmati ashram, CEPT University, Kankaria Lake, huthesseing temple among several others. This proves it to be a hotspot in Gujarat and tourist will be coming around the year to witness these wonders. Sabarmati ashram is the ashram which tells the life of Mahatma Gandhi. It is one of the most visited tourist places in Gujarat. This all adds jewel to the crown when it comes to investing in new flats in Ahmedabad.
How can a person not talk about education while talking about Ahmedabad? Ahmedabad is the host to several renowned institutes like the IIM-A which is the top management institute in whole of India. Universities like the CEPT University and Ahmedabad University quench the educational thirst of the city! CEPT university is Asia's one of the finest universities for architecture and civil engineering!
Thus, the above facts prove that investing in real estate is lot safer and would never fail. The growth and development of Gujarat as a state has also been remarkable and there should be no doubts for the Manchester of India. So invest in new flats in Ahmedabad or new projects in Ahmedabad to ensure that the money grows and the goddess of wealth resides in your flat!
What Is The Value Of A Good Landscape Designer?


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Perthshire Sport
New signing Craig Bryson keen to spark goal-scoring run at St Johnstone
The midfielder reminisced back to a time when former club Derby County were struggling for goals - until he netted a hat-trick
Matthew Gallagher
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Craig Bryson knows one moment in a game can turn around a team's fortunes in front of goal.
And the recently-signed St Johnstone midfielder is hoping to repeat a similar feat which helped his former club Derby County shake off a mini goal drought.
He fondly remembers netting a hat-trick for The Rams - they hadn't scored in four games - in a stunning 2014 English Championship victory against rivals Nottingham Forest.
Saints, who have netted only one goal in their last give games, are in need of a touch of inspiration to spark a productive spell in front of goal.
And Bryson said: "All it takes is one shot to go in from anyone and that turns things around for you.
"I remember at Derby we hadn't scored for four games before playing Nottingham Forest in a derby game.
"We went into it without a goal in a long time, but went out and beat them 5-0 that day.
"I got a hat-trick and the fans there will speak to me about it to this day.
"It was unbelievable, it's a huge game - you get 30,000-odd fans packing the stadium out.
"So that was a great memory from my time there and it just shows how things can pan out.
"We came off the back of four games without a goal thinking: 'We have Forest next, we need to get our finger out' and by half-time we were 3-0 up.
"After that game we went on a run and ended up in the play-off final. What you have to do is stay confident and keep doing the right things."
Bryson, most recently of Aberdeen, has been impressed with the Perth side's style of play this term.
He said: "If you watch the games we're getting into good positions and it will turn in our favour.
"As the gaffer said, if you are playing badly and losing you need to be worried - but that's not the case.
"We are playing good football and creating chances, it's just not going our way at the moment.
"It would be nice to bring goals because everyone has to contribute on that front.
"I have seen the team play a few times this season and the boys have been unlucky, they have played well but just not scored goals at the right times to go on and win.
"The Ross County game was like that, we dominated the game but just couldn't get a goal.
"I came off the bench in the Aberdeen game as well and that could have gone either way.
"From what I've heard about all the other games, that's happened a lot this season."
Saturday afternoon's trip to Livingston is set to be the 33-year-old's club debut after recovering from an upset stomach early in the week.
He told the PA: "I felt a bit ill at the start of the week with an upset stomach but with everything going on you don't want to come in and be around people.
"That's the first thing you think there days, you don't know what it is and what it could be.
"I'm just glad to get in and get going now, I can't wait to get playing. I'm fit and ready, raring to go after what happened last season.
"That was something new in my career, I've played a lot of games in my career and have never been out for that long before.
"We have got to the bottom of it now, the ankle injury was hurting other areas of my body because it was making me run differently.
"But thankfully now I am able to get out there pain-free. I want to push on now and get back to doing what I enjoy."
Bryson added: "I want to play with a smile on my face in the last few years of my career.
"Leaving Aberdeen was mutual on both parts, a lot of things came into play - the season I had, the pandemic and cost-cutting.
"It was the best thing for both of us."
Perth Sport
Scottish Professional Football LeagueGio van Bronckhorst and Rangers uninterested in Celtic postponement with NO derby delay approachA host of players could be missing for both sides but Van Bronckhorst was calm on the hot topic.
Matt O'Riley reveals crucial Celtic transfer phone call from Ange Postecoglou that instantly sold him on a move
Matt O'RileyThe midfielder has left MK Dons to pen a four-and-a-half year deal at Celtic Park.

redpajama_set_name: RedPajamaCommonCrawl


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Emgage Action Endorses Bernie Sanders
February 20, 2020 Deplorable Kel 2020 Elections, Breaking News, Islam in the U.S., Politics, U.S. Elections, U.S. News 0
Emgage Action has endorsed Bernie Sanders for President of the United States.
Emgage Action is a political organization with the goal of increasing the number of muslims serving in the United States government at all levels.
Over the last several years Emgage has endorsed many muslims such as Congresswomen Rashida Tlaib and Ilhan Omar, along with non-muslims who they feel will represent their muslim values.
Emgage announced their endorsement of Sanders on Twitter:
More than any other presidential candidate, Bernie Sanders has built an historically inclusive movement: one that is grounded in the belief that all humans are equal and worthy of a dignified life. We are proud to stand by his side. @BernieSanders
Bernie Sanders replied stating he was honored to receive the endorsement of Emgage Action:
It is an honor to receive the endorsement of @EmgageAction. While Donald Trump has demonized our Muslim communities, our movement is working to bring Muslims and people of all backgrounds into the political system.
Emgage Action is just one of many muslim political organizations seeking to increase the number of muslims in office. Unsurprisingly Emgage is not the first of these groups to endorse Sanders. The Muslim Caucus of America and The Muslim Democratic Club of New York both recently endorsed Sanders, as have most of the muslim candidates (past and present) running for office. It is more than clear that Sanders has the muslim vote.
By Deplorable Kel
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Emgage Action
The Muslim Democratic Club of New York Endorses Bernie Sanders
Muslim MN State Rep. Lashes Out At Republican Congressman For Saying Everyone Should Eat Pork


Jim Henderson came for a weekend to talk about leadership development, and in the midst of that he talked about 3Q.
Because it's scheduled, every single month after the first three Check-ins, they know I'm really interested in their development and in how they are doing in their role.
I'm finding when I checked in with Jim Henderson to show him how my Check-ins have gone — and the other team members were doing the same — he saw a pattern and said, "Are you aware that your other team members are dealing with the same issue that you're dealing with?" I thought it was a small problem with one of my people; it turned out it was a pervasive problem in my office — that I dealt with.
I think the opportunity I'm seizing the most business leadership development, not just leadership management. Before a problem comes up when I say, "How can I help?" I'm able to help my team solve that problem themselves.
I have been so much more conscientious and intentional about showing my team members how to be leaders. It's giving me space because of the calendar schedule, and it's giving me language to say, "This is how you lead." I give them the authority, I give them tools, I give them time — for them to talk out how they are leading and to see better ways for them to do it.
It's making me think more about how I'm leading, so I can say, "And this is how you can lead."
I see greater trust growing — more quickly — because of this intentionality. It's becoming cohesive because of those questions.
In the face of a competitive job offer, the 3Q system played a critical role in helping me to decide to stay put. I felt that my leadership was taking practical steps to understand, support and develop me as an individual and as a professional.
When I first started implementing the 3Q system with my own team, some of my staff said it felt formulaic and a little contrived... maybe awkward is a better word. After we'd been faithful in our meetings for several months, however, I missed a 3Q session with one of my grumblers only to receive the feedback that they felt like these 30 minutes were "my time with you... and it's important to me." That moment was a big one for me as a leader. I began to more deeply understand the importance of developing predictable rhythms when it comes to intentional time spent together.
3Q has allowed a set, monthly time with my direct supervisor to connect and touch base about challenges and successes and develop a stronger working relationship.
Meeting individually with each employee who reports to me has provided an opportunity to continue to get to know them as people. 3Q creates a safe environment to share their thoughts, concerns, and successes — it also gives me a glimpse of their perspective and how they process.
I don't have to worry about if/when I will get the chance to speak with my supervisor. I know that there are regularly scheduled 3Q meetings, and that any concern or question I have, I will have the opportunity to discuss with her at the next 3Q.
This is important because my supervisor is on the road most of the time, so I rarely see her. When she is on-site, she is swamped with meetings. Having time set aside for a 3Q, it allows me to have some face to face time with her, which may not feel "necessary" but in the grand scheme of things, it really is. The process helps me to feel valued.
I like how open-ended "How are you?" is. A few months ago, I was feeling rather disconnected with our other teammates because my boss travels, two team members are off-site, and the one who is on-site is rarely in the office. 3Q gave me a platform to share my sense of dissatisfaction. It wasn't a major complaint or an issue that would cause me to leave my job, but it was contributing to my feeling somewhat alone and forgotten about. By sharing these feelings with my supervisor, our team was able to come up with some workarounds to help me feel more connected and increase my satisfaction level on the job.
After just three weeks of 3Q check-ins, I had a dramatic drop in stress, solving problems I didn't know existed — be it a workplace relationship that wasn't effective, or an operational flaw that didn't have the right approach. Now I have more time. I stay on top of my workload by allowing and trusting my team who is actively, willingly helping me.
So many people can give you advice, but no one has ever in my life taught me how to put that advice into practice. The value I bring to my organization has grown exponentially by adding a little discipline and structure to the things I was already doing, but doing wrong. I cannot recommend this enough. 3Q has changed my life.
The structure and reliability of attention in 3Q has built invaluable trust between employees and leaders, which in turn has my team volunteering to take on more responsibility, making me more effective in my role.
When I was asked to start 3Q I didn't see how I could possibly have the time.The work was piling up, and the people who needed my attention where lined up out the door. I now plan everything on my schedule around my 3Q check-ins.
Definitely gained more overall awareness of the employees' thoughts/attitudes and potential fit in their roles. I have gained a better understanding of simply where we stand as a company. I have realized the need for a set of written standards in our company. I have learned more about how our employees think as opposed to just knowing how they perform. The employees love the one-on-one time, as do I.
I'm feeling a great sense of hope that our guys out in the field are going to find their place and thrive.
I am meeting with [our], HR Director, on February 14 to discuss how we integrate and infuse 3Q into the divsion's bloodstream through HR. This is our best hope to grow this beyond Major Kelly's personal passion and commitment to the process. Chris is currently rolling out our new performance appraisal process so we need to be careful to not overload folks, but we have discussed 3Q as a great tool that balances out the appraisal process. Such reviews tend to be set up where the person in power tells the person at a lower power level how well they are (or aren't doing). Goals are set but they are usually owned by the employee. While 3Q doesn't exactly reverse the process, it does provide a balance where the employee, in a sense, gives the power holder feedback on how well they are being led and developed. If these two balancing forms of communication can live together I think both will enhance the other.
"This [article from McKinsey & Company on The Fairness Factor in Performance Management] highlights the way 3Q functionally fits into a performance management system.
It sets the bar of fairness (a term I don't always like) by encouraging and facilitating regular (monthly) conversations where the power differential is leveled, if not reversed, giving an employee a voice and the listening ear of their supervisor.
I think this not only paves the way for the yearly performance conversation, but actually makes it natural. It is something talked about on a regular basis.
I have not started my performance conversations yet, but [one of my colleagues], who has, reports that she felt that much of what was discussed was already covered in 3Q and the one-on-ones that often follow it.
I am curious to see how this evolves. Thanks again."
— Divisional Head of HR for 501(c)(3) with 375 employees and several hundred of volunteers across 28 communities with an annual budget north of $6 million.


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Photos: Mike Groll (3), Frank Franklin II/APPhotos: Mike Groll (3), Frank Franklin II/AP
A Group of Democrats Joined Republicans to Give Them Power in New York. On Election Day, New Yorkers Wiped Them Out.
All eight members of Andrew Cuomo's IDC faced primary challenges on Election Day in New York. Most were wiped out.
September 13 2018, 6:22 p.m.
When New Yorkers go to the polls today, they'll have an opportunity to cast judgement on a group of Democratic state senators who, for years, caucused with Republicans, giving the GOP a constructive majority in a Democratic legislature. The group, known as the Independent Democratic Conference, or IDC, disbanded in April in a move that's been interpreted as an attempt to mitigate criticism ahead of primary challenges this fall. But all eight members of the IDC are nevertheless facing primaries. [See below for Thursday's results.]
One of the challengers is Julie Goldberg, a librarian from Chestnut Ridge who is hoping to unseat Sen. David Carlucci, D-Clarkstown. In an email to The Intercept, Goldberg described the IDC as an "unholy alliance." She said she decided to throw her hat in the ring when it became obvious that someone needed to challenge Carlucci's "betrayal." "Having no political ambitions," said Goldberg, "I had nothing to lose by challenging him."
Goldberg says that when she began her campaign, a network of over 60 progressive organizations had already been formed to oppose the IDC. The coalition, which includes No IDC NY, True Blue NY, the Working Families Party, and the New York Progressive Action Network, among many others, had been phone banking in IDC member districts since November 2017 with the goal of informing Democrats about what their state senators were up to. "By the time I declared, in May of this year, there was an organized team of volunteers from all over the district, in both Rockland and Westchester ready to transform itself into a campaign," said Goldberg.
That infrastructure helped Goldberg and other IDC challengers establish their campaigns. Most of those going up against IDC incumbents have been able to raise substantial sums and garner significant endorsements as a result of anti-IDC energy in the state. It's a blue wave, but one aimed at sitting Democrats. "All sorts of political players in the state are realigning," said Susan Kang, an associate professor of political science at John Jay College of Criminal Justice, and one of the co-founders of No IDC NY.
For years, the IDC effectively turned control of the state Senate over to the GOP, contrary to the wishes of voters who sent Democratic majorities to Albany in 2012 and 2016. The conference first came into existence in January 2011, when Sen. Jeff Klein, D-Bronx and three other Democrats — Carlucci; Diane Savino, D-Staten Island; and David Valesky, D-Oneida — broke from the Democratic caucus and joined what was then a Republican majority in exchange for committee chair positions and increased power. "They decided that they would make alliance with the Republicans to be in power and get perks," said George Albro, co-chair of the New York Progressive Action Network.
Sen. Jeffrey Klein speaks during a news conference on Jan. 5, 2011 in Albany, N.Y., where he announced that he and three other Democratic senators were forming an independent Democratic conference. Joining Klein, from left, are Sen. Diane Savino, Sen. David Carlucci and Sen. David Valesky.
Photo: Mike Groll/AP
A year later, in 2012, the Democrats won a majority of seats in the chamber. But Klein, Carlucci, Savino, and Valesky, joined by then-Sen. Malcolm Smith, D-Queens, decided to stay aligned with the GOP senators. "They gave what would otherwise be a minority party control against their own party," said Albro.
Smith was removed from the caucus after his indictment on corruption charges in 2013, but the caucus picked up Tony Avella, D-Queens, the next February. Avella and Klein faced primary challenges in 2014, but both survived. The rest of the conference members were unchallenged in their primaries. In 2016, after the general election, Sens. Marisol Alcantara, D-Manhattan, and Jesse Hamilton, D-Brooklyn, joined the IDC; in January 2017, Sen. Jose Peralta, D-Queens, became a member.
That's where things stood until April of this year. The group dissolved following a meeting with Gov. Andrew Cuomo, who has long been seen as an ally and quiet backer of a Republican-led state Senate. "It's clear that Cuomo was behind this," said Albro. "He supported and never challenged them, then this year he called them into his office and they immediately disbanded."
The Intercept reached out to each IDC member's campaign for comment, but never received a reply. Cuomo's office did not return a request for comment, but in the past, the office has denied that the governor plays any role in the IDC. Cuomo's comments on the conference are few and far between, but he did suggest in May 2017 that he preferred a Republican Senate to a Democratic one, if only for the efficacy of the former caucus in Albany.
"We've had a unified Democratic government in Albany," Cuomo told the Daily News, referring to the 2009-10 session when Democrats controlled the Senate, Assembly, and governor's mansion, and when Cuomo was the state attorney general. "It's not a hypothetical. We've had it. It wasn't extraordinarily successful. So I work with the Assembly and Senate that I've been given, and I do the best I can."
The IDC has come under repeated criticism for stymying the ability of Democrats in the Senate to get things done. Progressives cite the caucus's alliance with Republicans as one of the factors preventing the Senate from addressing legislation — like single-payer health care, marijuana legalization, and the DREAM Act — meant to improve the lives of New Yorkers. Cuomo didn't seem interested in exercising his power over what could pass through the Senate, Kang told The Intercept. "Cuomo made sure not much reform got through anyway."
Now, on election day, the moment of reckoning appears to be upon the IDC. Though no public polling has been done on the individual races, there are indications that at least some of the caucus members won't be returning to Albany. The Senate Independence Campaign Committee, or SICC, the IDC's campaign fundraising account, is funneling money to five of the incumbents: Alcantara, Hamilton, Klein, Avella, and Peralta.
A rundown of all of Thursday's IDC races follows.
This Queens race is a rematch between former comptroller John Liu and Avella, who Liu first challenged in 2014. Liu thinks he's got him this time. He's raised more money than Avella, and has the endorsement of both of the district's congressional representatives and both of its assembly members. Liu made a late-game decision to run in July, meaning he had to get 3,000 signatures in the 10 days before the registration deadline. Liu sees his ability to accomplish that feat as evidence of strong voter support for his challenge. "That's how much enthusiasm there is for beating the IDC," said Kang.
A former aide to Mayor Bill de Blasio, Jessica Ramos is running to unseat Peralta in Queens. Ramos has the support of the mayor and has also been endorsed by the New York Times, Cynthia Nixon, and U.S. Sen. Kirsten Gillibrand, as well as a number of progressive groups, including Our Revolution and People for Bernie. Notably, Gillibrand may be looking to absolve herself for her prior endorsement of Rep. Joe Crowley in his race against Alexandria Ocasio-Cortez. (Ramos also endorsed Crowley).
Like Liu, Ramos has also out-earned Peralta, with $108,065.29 on hand as of September 11, compared to Peralta's $101,114.35.
Zellnor Myrie, a lawyer, is going up against Hamilton in the 20th District, which encompasses parts of northern Brooklyn. Myrie, like Julia Salazar, is attacking his opponent's ties to the real estate industry and questionable progressive credentials. Myrie, like Ramos, has been endorsed by the Times, Gillibrand, and de Blasio, among others.
23rd District
Staten Island's Savino is up against Jasmine Robinson. Savino has significantly out-raised and outspent her opponent: Robinson's July campaign filing shows a paltry $4,442.23, compared to Savino's $141,777.90.
Despite endorsements from progressive groups like Our Revolution and Citizen Action, Robinson may be outmatched by Savino's machine. And Robinson's heart may not be in the race: She filed suit in July to be taken off the ballot due to fraudulent signatures on her own nomination papers, though she quickly reversed herself.
No IDC NY's chief strategist Gus Christensen believes that Robinson was taken advantage of by a political team that didn't have her best interests at heart. "She almost got knocked out," said Christensen. "But she got right back up; she got a new team who have her best interests at heart."
31st District
With more money on hand than Alcantara, despite her outspending him, challenger Robert Jackson could be poised to upset the Manhattan incumbent. Alcantara was a recent arrival to the state Senate, while Jackson, a former city councilor, has a long history in the district. Jackson's efforts in the district include his work on public education: He co-founded the Campaign for Fiscal Equity, which sued the state in 1991 for not adequately providing funding to the city's schoolchildren. The campaign finally won the lawsuit in 2006, but funding has still been hard to come by due to inaction by the state legislature and Cuomo.
Like the other candidates, Jackson can claim a long list of endorsements, including former Mayor David Dinkins and U.S. Rep. Jerrold Nadler.
"He's running to help us overcome the biggest obstacle to progressive change here in New York," said Jackson's spokesperson Richard Fife, "an obstacle that's been around for decades: the Republican-controlled state Senate."
In the Bronx, Klein, the IDC's leader, is facing off against political newcomer Alessandra Biaggi. Biaggi is no stranger to Democratic politics: She was Hillary Clinton's deputy national operations director in 2016, and her grandfather, Mario Biaggi, was the congressional representative for New York's 16th District from 1969 to 1988, when he resigned following corruption charges.
Biaggi's challenge has made an impact on Klein: He has spent $836,172.64 in the last 30 days, including a $124,035.90 check to the Hamilton Campaign Network, a firm run by actor Lin-Manuel Miranda's father, for "Digital Ad Production and Placement." Meanwhile, Biaggi has been knocking doors in the district with Ocasio-Cortez, the reigning queen of Bronx upsets. Klein has a massive cash advantage over Biaggi, with $957,827.45 to her $263,113.17, but his significant spending, over $2.4 million in all, suggests he's nervous.
A victory over Klein would not only bolster Biaggi's profile, but defeating the IDC leader would be a major blow to the IDC and Cuomo (should Cuomo fend off the challenge from Nixon).
Carlucci is facing a challenge from Goldberg in Westchester. Goldberg got into the race late, and the 38th District presents unique challenges for an upstart campaign. The district is geographically spread out, making door-to-door canvassing more difficult. And Goldberg is hurt by her relative lack of personal political connections. "She has much less local political support from other elected officials than our candidates inside of New York City do," said Christensen, "and it is that much harder for her to get the kind of media attention that would help build her fundraising base." Carlucci has not recently received SICC money, but he still out-raised Goldberg by a factor of 24. This financial advantage suggests he is less vulnerable than some of his opponents, but as Ocasio-Cortez's win showed, money isn't everything.
In New York's northernmost district, the IDC's former deputy leader, Valesky, is facing a challenge from Rachel May, a sustainability education coordinator at Syracuse University. May's progressive campaign — she supports single-payer health care and campaign finance reform — could turn out voters on its own merits. But Valesky has outspent May by nearly 800 percent — $286,118.87 to her $35,847.32. Still, May believes she has a shot, citing her endorsement by Valesky's own Oneida County Democratic Committee as an indication that her support is stronger than it appears.
The majority of members of the Independent Democratic Conference, or IDC, have been defeated in the New York state and local primary election. Six members of the defunct group, which broke away from Senate Democrats to hand control of the chamber to the Republicans in 2011 and caucused with the GOP until April, followed their leader, Jeffrey Klein, in defeat. Klein fell to Alessandra Biaggi in a shocking upset. It's a steep fall from political grace for Klein, who as the leader of the IDC was one of the most powerful people in Albany for nearly seven years, and leaves just two of the former IDC members behind.
In Manhattan, Robert Jackson, a former city councilor, who ran explicitly on defeating the IDC, beat Marisol Alcantara. Zellnor Myrie, a lawyer and activist, defeated incumbent Jesse Hamilton in the 20th. Former Mayor Bill de Blasio aide Jessica Ramos unseated incumbent Jose Peralta in the 13th. And in the 11th, former City Comptroller John Liu beat incumbent Tony Avella in Liu's second challenge to the sitting senator.
The biggest upset of the night, however, goes to Syracuse University administrator Rachel May in Oneida County, who appears to have narrowly beaten David Valesky. Incumbent Valesky outspent May by some 800 percent but May was able to defeat the incumbent, despite lacking the geographical advantage of the other victors, who all were centered in New York.
In the IDC's column, only two incumbents are likely going back to Albany. Diane Savino handily beat back a somewhat disorganized challenge from Jasmine Robinson in Staten Island and David Carlucci defeated Julie Goldberg in Rockland County's 38th.
The IDC's power to regroup after the November elections and determine the makeup of the next Senate has effectively been destroyed. The insurgent victories against the conference represent a minor victory in the latest battle in the Democratic Party's civil war, which was offset by Gov. Andrew Cuomo's decisive win against Cynthia Nixon.
State Sen. Martin Dilan, meanwhile, was defeated by democratic socialist Julia Salazar, despite a run of high-profile scandals in the weeks leading up to the election. Though not IDC-connected, Dilan was associated with the group after receiving funding from the IDC's campaign finance fundraising organization, the Senate Independence Campaign Committee, or SICC, in 2016. On the other side of related results, Simcha Felder, a Democrat who caucuses with Republicans — though not with the IDC — and is known as one of the most right-wing members of the Senate, easily beat a challenge from lawyer Blake Morris.
New York Working Families Party Director Bill Lipton declared victory. "New York politics changed forever tonight," he said. "The IDC is dead. The center of gravity has shifted, and Andrew Cuomo will face a radically different Albany. For years, Cuomo, the IDC and the Republicans led a government which blocked countless progressive policies. Now, the WFP and a progressive insurgency has ended the IDC, and Cuomo will have to deal with something new for him after November: a Democratic legislature."
Polls closed at 9 p.m., but voters reported throughout the day on social media about irregularities with respect to registration and proper documentation. Many voters posted about having to fill out affidavits in order to vote and having their names in the wrong lines. Journalist Sean T. Collins collected many of the stories in a Twitter thread.
Vote suppression in @NYGovCuomo's New York: a thread.
— Sean T. Collins (@theseantcollins) September 13, 2018
Top photo: From left, Jesse Hamilton, Jeffrey Klein, David Carlucci, and Tony Avella.
Eoin Higgins[email protected]gmail.com@EoinHiggins_
Andrew Cuomo Keeps Hiring Republicans
Zephyr Teachout's Fundraising Spikes In Wake of New York Times Endorsement in Attorney General Race
EMILY's List Backs Letitia James, Andrew Cuomo's Candidate for New York Attorney General


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redpajama_set_name: RedPajamaGithub


In Spring 2007, Dan Halutz, former chief of staff of the Israeli military, participated in an exclusive 8 week managmeent course at Harvard Business School. Halutz oversaw the 2006 Lebanon war during which he ordered Israeli pilots to bomb 10 multistory buildings in the Dahiya district of Beirut as a response to Hezbollah's rocket fire on Haifa.
The Harvard student organization Alliance for Justice in the Middle East organized a campaign and action to bring attention to the university's relationship with people accused of war crimes. Recently released Wikileaks cables include an Israeli media summary from the U.S. Embassy in Tel Aviv from May 2007 which discusses the Harvard student action.
Maariv reported that a far-left wing Harvard University group that includes Jews and Israelis distributed "wanted for crimes of war" posters for former IDF chief of staff Dan Halutz.
I don't doubt that the war criminals in Israel and the US government paid attention to the Harvard protest. What concerns me is: what are they planning to do about it?
Hopefully, Harvard (from the trustees--admin.--students) will join the boycott of Israeli products.
Sometimes universities listen. I was in student demonstrations at UC Davis during the Cambodian invasion. The administration let us look at their files to see if the campus had war-related research, which it did. It isn't always easy, though. YOu have to struggle for it. Good luck to the protesters at Harvard!


Guest Post: Can Creativity Be Measured in a Laboratory?
Dr. Mark Changizi, a cognitive science researcher, and professor at Rensselaer Polytechnic Institute, is one of the most exciting rising stars of science writing and the neurobiology of popular culture phenomena. His latest book, The Vision Revolution, expounds on the evolution and nuances of the human eye—a meticulously designed, highly precise technological marvel that allows us to have superhuman powers. You heard me right; superhuman! X-ray vision, color telepathy, spirit reading, and even seeing into the future. Dr. Changizi spoke about these ideas, and how they might be applied to everything from sports stars with great hand-eye coordination to modern reading and typeface design with us in ScriptPhD.com's inaugural audio podcast. He also provides an exclusive teaser for his next book with a guest post on the surprising mindset that makes for creative people. Read Dr. Changizi's guest post and listen to the podcast under the "continue reading" cut.
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text:
A LOUTH TRAGEDY . . . Part One
Back in the 1950s and 60s I was a regular visitor to Louth. Mother was a Louth girl, and my grandmother, although born in Yorkshire, had lived in the area since before The Great War, first in a cottage in the grounds of Tathwell Hall, where her father was Head Groom, and then in the town itself. I used to stay with her in Tennyson Road during school holidays, and blissful days they were. I had made several friends in Louth, and we did what most lads did in those innocent days – played football and cricket and went fishing. The most intense of all our passions was, however, trainspotting.
Each summer, we used to buy what was called a Runabout Ticket. It cost twelve shillings and sixpence and was a small rectangle of stiff blue cardboard. On it was printed a stylized map of the railway network in the area. It meant we could travel on any train, between any of the stations, as often as we wanted, for a week. Crucially, it gave us access to the East Coast main line between Peterborough and Grantham, with its magnificent Gresley Pacifics and all manner of spectacular engines. At other times, however, we had to make do with the trains that ran through Louth on the Grimsby to Boston line, and one of our favourite places to trainspot was the level crossing on Stewton Lane. Between trains.we could muck about by the nearby stream where it ran through a little gully, which was known as Seven Trees Island. Happily undreamed of in those days was the fact that we were enjoying ourselves on the site of one of the most gruesome and tragic murder cases of the early twentieth century.
In 1927, just by the railway, there stood a small wooden bungalow, the home of Bertram Horace Kirby, 46, and his wife Minnie. They were not originally Louth people, both having come from Boston, where they had married – possibly in St Botolph's – in 1905. They had three sons. The two oldest had left home. Harry, 21, lodged with Mrs Took in nearby Church Street, while Ralph, 17, had emigrated to Canada. There was a much later addition to the family. Leslie Norman Kirby was just 8 years old. Minnie Eleanor Kirby (right) was described as follows in subsequent press reports:
"Mrs. Kirby was tall, and of striking appearance. She was most friendly woman, and was liked very much by her neighbours. Her hobby was gardening. She had studied her subject, and she was an expert gardener, and passionately fond of her flowers. She was extremely well-read, and was a thoroughly cultured woman, clever in many ways, and musical. She was a keen churchwoman.
Mrs. Kirby was also an enthusiastic member of the Women's Unionist Association, and canvassed in Louth at the last election. She had been educated in Boston at Miss Stothert's High School, and for about 10 years she was head clerk at Mr. A. Simpson's furniture store, in the Market-place. She was very well-known in Boston. At school, she was a very apt pupil, and we are informed that Miss Stothert "thought a lot of her." During her schooldays the took part in "The Mandarin", and other plays. She was extremely fond of rowing, and frequently enjoyed her favourite exercise on the Witham."
Bertram Horace Kirby (left) was a year younger than his wife. He also had musical talent, and while they lived in Boston he had been church organist in the town, and the village of Frampton. He had applied himself to various trades while living in the Boston area, but had worked for almost ten years for the London and North Eastern Railway. By 1927, however, he had given this job up, and had attempted to strike out on his own as a commercial dealer.
In the early evening of Sunday 10th July , Harry Kirby called at his parents' home, and took his mother for a stroll. Although he had moved out of the bungalow because he and his father "couldn't get along" he had noticed nothing untoward in the atmosphere between his mother and father. He was later to admit that his father "was prone to violent tempers when he had taken drink."
Despite all seeming well with his parents, Harry Kirby must have had a sixth sense that prompted him to visit the bungalow on Tuesday 12th July. He found the doors locked and the curtains drawn, and could make nobody hear. He walked into the town and asked for help from the police. He returned with Inspector Davies who must have also sensed something was wrong, and forced his way into the bungalow through a rear window. What he found confirmed Harry Kirby's fears that a tragedy was about to unfold. Minnie Kirby was lying dead on the floor of the living room, on her stomach, with her head turned to one side. At the base of her skull was a savage wound which had almost separated her head from her body.
In Part Two
AN ARREST INTERRUPTS A GAME OF DOMINOES
ANGUISHED LETTERS


text:
\section{Introduction}

Let $k$ be a commutative artinian ring and $\check k$ be the minimal injective cogenerator. We denote by $D=\Hom_k(-,\check k)$ the Matlis duality. The categories we consider are $k$-linear Hom-finite Krull-Schmidt and skeletally small.

Recall from \cite{LenzingZuazua2004Auslander} that an abelian category $\CA$ is said to \emph{have Auslander-Reiten duality}, if for any objects $X$ and $Y$, there exist natural isomorphisms
\[
  \overline\Hom_\CA(Y,\tau X)
  \simeq D\Ext_\CA^1(X,Y)
  \simeq \underline\Hom_\CA(\tau^-Y,X).
\]
Here $\tau$ and $\tau^-$ are mutually quasi-inverse equivalences between the stable categories of $\CA$. More generally, the notion of \emph{generalized Auslander-Reiten duality} for an exact category $\CC$ was introduced in \cite[Section~3]{JiaoGeneralized}. We denote by $\CC_r$ the smallest additive subcategory of $\CC$ containing projectives and indecomposable objects which appear as the third term of an almost split conflation, and by $\CC_l$ the smallest additive subcategory containing injectives and indecomposable objects which appear as the first term of an almost split conflation. We still have the mutually quasi-inverse equivalences $\tau$ and $\tau^-$ between stable categories of $\CC_r$ and $\CC_l$.

Auslander introduced the notion of \emph{morphisms determined by objects} in his Philadelphia note \cite{Auslander1978Functors}, which extends Auslander-Reiten theory in some aspects. We study morphisms determined by objects in an exact category $\CC$. We prove that a deflation in $\CC$ is right determined by some object if and only if its intrinsic kernel lies in $\mathcal{C}_l$; see Theorem~\ref{thm:det}. We give characterizations for an object lying in $\CC_r$ via morphisms determined by objects; see Theorem~\ref{thm:C}.

Following \cite{Krause2013Morphisms} and \cite{ChenLe2015note}, we introduce the notions of \emph{having right stably determined deflations} and \emph{having left stably determined inflations}. We show that the conditions ``$\CC$ has Auslander-Reiten duality'', ``$\CC$ has right stably determined deflations'' and ``$\CC$ has left stably determined inflations'' are equivalent; see Theorem~\ref{thm:AR}.

The paper is organized as follows. In Sections~2 we recall some basic properties of morphisms determined by objects. In Section~3 we prove that if  two objects $C$ and $C'$ are isomorphic in the projectively stable category, then a deflation is right determined by $C$ if and only if it is right determined by $C'$. In Section~4 we prove the following existence theorem: given objects $C\in\CC_r$ and $Y\in\CC$, for certain submodules $H$ of $\Homc(C,Y)$, there exists a deflation $\alpha\colon X\to Y$ with $H=\Im\Homc(C,\alpha)$. Sections~5, 6 and 7 are dedicated to the proofs of Theorems~\ref{thm:det}, \ref{thm:C} and \ref{thm:AR}, respectively.

\section{Morphisms determined by objects}

Let $\CC$ be an additive category. We recall some well-known notions. Here, we use the terminologies in \cite[Section~3]{Ringel2013Auslander}.
Let $f\colon X\to Y$ be a morphism and $C$ be an object. We call $f$ \emph{right $C$-determined} (or right determined by $C$) and call $C$ a \emph{right determiner} of $f$, if the following condition is satisfied: each morphism $g\colon T\to Y$ factors through $f$, provided that for each $h\colon C\to T$ the morphism $g\circ h$ factors through $f$. If moreover $C$ is a direct summand of any right determiner of $f$, we call $C$ a \emph{minimal right determiner} of $f$.

\begin{lemma}\label{lem:det.sum}
  Let $f_1\colon X_1\to Y_1$ and $f_2\colon X_2\to Y_2$ be two morphisms. Then for an object $C$, the morphism $\smatrix{f_1 &0\\ 0 &f_2}\colon X_1\oplus X_2\to Y_1\oplus Y_2$ is right $C$-determined if and only if both $f_1$ and $f_2$ are right $C$-determined.
\end{lemma}

\begin{proof}
  For the sufficiency, we assume that $f_1$ and $f_2$ are right $C$-determined. Let $\smatrix{g_1\\ g_2}\colon T\to Y_1\oplus Y_2$ be a morphism such that for each $h\colon C\to T$, there exists some morphism $\smatrix{s_1\\ s_2}\colon C\to X_1\oplus X_2$ satisfying $\smatrix{g_1\\ g_2} h=\smatrix{f_1 &0\\ 0 &f_2} \smatrix{s_1\\ s_2}$. We obtain $g_1\circ h=f_1\circ s_1$ and $g_2\circ h=f_2\circ s_2$. Since $f_1$ and $f_2$ are right $C$-determined, there exist some morphisms $t_1\colon T\to X_1$ and $t_2\colon T\to X_2$ such that $g_1=f_1\circ t_1$ and $g_2=f_2\circ t_2$. The morphism $\smatrix{t_1\\t_2}\colon T\to X_1\oplus X_2$ satisfies $\smatrix{g_1\\ g_2} = \smatrix{f_1 &0\\ 0 &f_2} \smatrix{t_1\\ t_2}$. It follows that $\smatrix{f_1 &0\\ 0 &f_2}$ is right $C$-determined.

  For the necessity, we assume that $\smatrix{f_1 &0\\ 0 &f_2}$ is right $C$-determined. Let $g\colon T\to Y_1$ be a morphism such that for each $h\colon C\to T$, there exists some morphism $s\colon C\to X_1$ satisfying $g\circ h=f_1\circ s$. The morphisms $\smatrix{g\\0}\colon T\to Y_1\oplus Y_2$ and $\smatrix{s\\0}\colon C\to X_1\oplus X_2$ satisfy $\smatrix{g\\ 0} h=\smatrix{f_1 &0\\ 0 &f_2} \smatrix{s\\ 0}$. Since $\smatrix{f_1 &0\\ 0 &f_2}$ is right $C$-determined, there exists some morphism $\smatrix{t_1\\ t_2}\colon T\to X_1\oplus X_2$ such that $\smatrix{g\\ 0} = \smatrix{f_1 &0\\ 0 &f_2} \smatrix{t_1\\ t_2}$. We obtain $g=f_1\circ t_1$. It follows that $f_1$ is right $C$-determined. Similarly, the morphism $f_2$ is right $C$-determined.
\end{proof}

\begin{lemma}\label{lem:det.PB}
  Let $f\colon X\to Y$ and $g\colon Z\to Y$ be two morphisms. Assume that $f'\colon E\to Z$ and $g'\colon E\to X$ form the pullback of $f$ and $g$. If $f$ is right $C$-determined for some object $C$, then $f'$ is also right $C$-determined.
\end{lemma}

\begin{proof}
  Let $h\colon T\to Z$ be a morphism such that for each $u\colon C\to T$, there exists some morphism $v\colon C\to E$ satisfying $h\circ u=f'\circ v$. Then we have $g\circ h\circ u=g\circ f'\circ v=f\circ g'\circ v$.
  Since $f$ is right $C$-determined, there exists some morphism $s\colon T\to X$ such that $g\circ h=f\circ s$. Since $f'$ and $g'$ form the pullback of $f$ and $g$, there exists a morphism $t\colon T\to E$ such that $h=f'\circ t$. It follows that $f'$ is right $C$-determined.
\end{proof}

Let $f\colon X\to Y$ be a morphism. Recall that $f$ is called \emph{right almost split} if $f$ is not a retraction and each morphism $g\colon Z\to Y$ which is not a retraction factors through $f$. Dually, we call $f$ \emph{left almost split} if $f$ is not a section and each morphism $g\colon X\to Z$ which is not a section factors through $f$.

Given two objects $X$ and $Y$, we denote by $\radc(X,Y)$ the set of morphisms $f\colon X\to Y$, that for any object $Z$ and any morphisms $g\colon Z\to X$ and $h\colon Y\to Z$, the morphism $h\circ f\circ g$ lies in $\rad\Endc(Z)$. Then $\radc$ forms an ideal of $\CC$. We observe by \cite[Corollary~2.10]{Krause2015Krull} that
\begin{equation}\label{eq:rad}
  \radc(X,Y) = \set{ f \colon X \to Y \middle|
    f\circ g\in\rad\Endc(Y),
    \mbox{ for each }
    g\colon Y\to X}.
\end{equation}

A morphism $g\colon Z\to Y$ is said to \emph{almost factor through} $f\colon X\to Y$, if $g$ does not factor through $f$, and for each object $T$ and each morphism $h\in\radc(T,Z)$, the morphism $g \circ h$ factors through $f$.

\begin{proposition}\label{prop:ras}
  Let $f\colon X\to Y$ and $g\colon Z\to Y$ be two morphisms, such that $\Endc(Z)$ is a local ring and $g$ almost factors through $f$. Assume that $f'\colon E\to Z$ is the pullback of $f$ along $g$. Then $f'$ is right almost split.
\end{proposition}

\begin{proof}
  We observe that $f'$ is not a retraction, since $g$ does not factors through $f$. Given an object $T$, assume that $h\colon T\to Z$ is not a retraction. We observe by (\ref{eq:rad}) that $h\in\radc(T,Z)$, since $\Endc(Z)$ is local. Since $g$ almost factors through $f$, there exists some morphism $s\colon T\to X$ such that $g\circ h = f\circ s$. By the pullback diagram, there exists a morphism $t\colon T\to E$ such that $h=f'\circ t$. It follows that $f'$ is right almost split.
\end{proof}

\section{Deflations determined by objects}

Let $k$ be a commutative artinian ring. We denote by $k$-mod the category of finitely generated $k$-modules. Let $\check{k}$ be the minimal injective cogenerator for $k$. Then the Matlis duality $D=\Hom_k(-,\check{k})$ is a self-duality of $k$-mod. From now on, the categories we consider are $k$-linear Hom-finite and Krull-Schmidt.

Let $\CC$ be an exact category. Recall that an \emph{exact category} is an additive category $\CC$ together with a collection $\CE$ of kernel-cokernel pairs which satisfies the axioms in \cite[Appendix~A]{Keller1990Chain}. Here, a kernel-cokernel pair means a sequence of morphisms $X\xrightarrow{i}E\xrightarrow{d}Y$, which we denote by $(i,d)$, such that $i$ is the kernel of $d$, and $d$ is the cokernel of $i$. Each kernel-cokernel pair $(i,d)\in\CE$ is called a \emph{conflation}, while $i$ is called an \emph{inflation} and $d$ is called a \emph{deflation}.
Given a conflation $\eta\colon X\to E\to Y$, for each $f\colon Z\to Y$ we denote by $\eta.f$ the conflation $\Extc(f,X)(\eta)$; for each $g\colon X\to Z$ we denote by $g.\eta$ the conflation $\Extc(Y,g)(\eta)$.

Let $\alpha_1\colon X_1\to Y_1$ and $\alpha_2\colon X_2\to Y_2$ be two morphisms. We mention the fact that $\smatrix{\alpha_1 &0\\ 0 &\alpha_2}\colon X_1\oplus X_2\to Y_1\oplus Y_2$ is a deflation if and only if both $\alpha_1$ and $\alpha_2$ are deflations;
see \cite[Proposition~2.9 and Corollary~2.18]{Buhler2010Exact}.

Recall from \cite[Section~2]{LenzingZuazua2004Auslander} that a morphism $f\colon X\to Y$ is called \emph{projectively trivial} if for each $Z$, the induced map $\Extc(f,Z)\colon \Extc(Y,Z)\to \Extc(X,Z)$ is zero. We observe that $f$ is projectively trivial if and only if $f$ factors through each deflation ending at $Y$. Dually, we call $f$ \emph{injectively trivial} if for each $Z$, the induced map $\Extc(Z,f)\colon \Extc(Z,X)\to \Extc(Z,Y)$ is zero.

Given two objects $X$ and $Y$, we denote by $\CP(X,Y)$ the set of projectively trivial morphisms from $X$ to $Y$. Then $\CP$ forms an ideal of $\CC$. The \emph{projectively stable category} $\underline\CC$ of $\CC$ is the factor category $\CC/\CP$. Given a morphism $f\colon X\to Y$, we denote by $\underline f$ its image in $\underline\CC$. We denote by $\Homcu(X,Y) = \Homc(X,Y)/\CP(X,Y)$ the set of morphisms from $X$ to $Y$ in $\underline\CC$.

Dually, we denote by $\CI(X,Y)$ the set of injectively trivial morphisms from $X$ to $Y$. The \emph{injectively stable category} $\overline\CC$ of $\CC$ is the factor category $\CC/\CI$. Given a morphism $f\colon X\to Y$, we denote by $\overline f$ its image in $\overline\CC$. We denote by $\Homco(X,Y)=\Homc(X,Y)/\CI(X,Y)$ the set of morphisms from $X$ to $Y$ in $\overline\CC$.

\begin{lemma}\label{lem:ft.stable}
  Let $\alpha\colon X \to Y$ be a deflation. Then a morphism $f\colon T\to Y$ factors through $\alpha$ in $\CC$ if and only if $\underline f$ factors through $\underline\alpha$ in $\underline\CC$.
\end{lemma}

\begin{proof}
  It is sufficient to show the sufficiency. Assume that $\underline f$ factors through $\underline\alpha$ in $\underline\CC$. Then there exists some morphism $g\colon T\to X$ in $\CC$ such that $\underline f = \underline\alpha\circ\underline g$ in $\underline\CC$. We have that $f-\alpha\circ g$ is projectively trivial in $\CC$. Since $\alpha$ is a deflation, there exists some morphism $h\colon T\to X$ such that $f-\alpha\circ g=\alpha\circ h$ in $\CC$. It follows that $f=\alpha\circ(g+h)$, factoring through $\alpha$ in $\CC$.
\end{proof}

\begin{lemma}\label{lem:det.stable}
  Let $C$ be an object. Then a deflation $\alpha\colon X \to Y$ is right $C$-determined in $\CC$ if and only if $\underline\alpha$ is right $C$-determined in $\underline\CC$.
\end{lemma}

\begin{proof}
  For the sufficiency, we assume that $\underline\alpha$ is right $C$-determined in $\underline\CC$. Let $f\colon T\to Y$ be a morphism in $\CC$ such that for each $g\colon C\to T$, the morphism $f\circ g$ factors through $\alpha$ in $\CC$. Then $\underline f \circ \underline g$ factors through $\underline\alpha$ in $\underline\CC$. Since $\underline\alpha$ is right $C$-determined in $\underline\CC$, we have that $\underline f$ factors through $\underline\alpha$ in $\underline\CC$. By Lemma~\ref{lem:ft.stable}, the morphism $f$ factors through $\alpha$ in $\CC$. It follows that $\alpha$ is right $C$-determined in $\CC$.

  For the necessity, we assume that $\alpha$ is right $C$-determined in $\CC$. Let $f\colon T\to Y$ be a morphism in $\CC$ such that for each $g\colon C\to T$, the morphism $\underline{f} \circ \underline{g}$ factors through $\underline\alpha$ in $\underline\CC$. By Lemma~\ref{lem:ft.stable}, the morphism $f\circ g$ factors through $\alpha$ in $\CC$. Since $\alpha$ is right $C$-determined in $\CC$, we have that $f$ factors through $\alpha$ in $\CC$. Then $\underline f$ factors through $\underline\alpha$ in $\underline\CC$. It follows that $\underline\alpha$ is right $C$-determined in $\underline\CC$.
\end{proof}

\begin{proposition}\label{prop:det}
  Let $C$ and $C'$ be two objects. Assume that $C\simeq C'$ in $\underline\CC$. Then a deflation $\alpha\colon X \to Y$ is right $C$-determined if and only if it is right $C'$-determined.
\end{proposition}

\begin{proof}
  We observe that $\underline\alpha$ is right $C$-determined in $\underline\CC$ if and only if $\underline\alpha$ is right $C'$-determined in $\underline\CC$. Then the result follows by applying Lemma~\ref{lem:det.stable} twice.
\end{proof}

\section{An existence theorem}

Let $\CC$ be a Hom-finite Krull-Schmidt exact category. Given an object $C$, we denote by $\Gamma_C=\Endc(C)^{\op}$ the opposite algebra of the endomorphism algebra of $C$, and by $\add C$ the category of direct summands of finite direct sums of $C$.

Recall from \cite[Section~2]{JiaoGeneralized} two full subcategories $\CC_r$ and $\CC_l$ of $\CC$ as follows:
\[
  \CC_r = \set{X\in\CC \middle| \mbox{the functor }
  D\Extc(X,-) \colon \overline\CC \to k
  \mbox{-mod is representable}}
\]
and
\[
  \CC_l = \set{X\in\CC \middle| \mbox{the functor }
  D\Extc(-,X) \colon \underline\CC \to k
  \mbox{-mod is representable}}.
\]
Then we have the mutually quasi-inverse equivalences
\[
  \tau \colon \underline{\CC_r}
  \overset\sim\longrightarrow
  \overline{\CC_l}
  \mbox{ and }
  \tau^- \colon \overline{\CC_l}
  \overset\sim\longrightarrow
  \underline{\CC_r}.
\]
For each $X\in\CC_r$, we have a natural isomorphism
\[\Homco(-,\tau X)  \overset\sim\longrightarrow D\Extc(X,-);\]
for each $X\in\CC_l$, we have a natural isomorphism
\[\Homcu(\tau^-X,-) \overset\sim\longrightarrow D\Extc(-,X).\]

Recall that a conflation $\eta\colon X \xrightarrow{u} E \xrightarrow{v} Y$ is \emph{almost split} if $u$ is left almost split and $v$ is right almost split. Assume that $\eta$ is almost split. We mention that both $X$ and $Y$ are indecomposable. By the following lemma, which is due to \cite[Proposition~2.4]{JiaoGeneralized}, we have $X\in\CC_l$ and $Y\in\CC_r$. We mention that $Y\simeq\tau^-X$ in $\underline\CC$ and $X\simeq\tau Y$ in $\overline\CC$; see the proof of \cite[Lemma~3.2]{JiaoGeneralized}.

\begin{lemma}\label{lem:ass}
    Let $X$ be an indecomposable object.
  \begin{enumerate}
    \item\label{item:lem:X:1}
      Assume that $X$ is non-projective. Then $X\in\CC_r$ if and only if there exists an almost split conflation ending at $X$.
    \item\label{item:lem:X:2}
      Assume that $X$ is non-injective. Then $X\in\CC_l$ if and only if there exists an almost split conflation starting at $X$.
    \qed
  \end{enumerate}
\end{lemma}

The classical case of the following lemma is well known; see \cite[Corollary~3.5]{Ringel2013Auslander}.

\begin{lemma}\label{lem:ker.det}
  Let $\alpha\colon X\to Y$ be a deflation with $\Ker\alpha\in\CC_l$. Then $\alpha$ is right $\tau^-(\Ker\alpha)$-determined.
\end{lemma}

\begin{proof}
  Let $f\colon T\to Y$ be a morphism such that for each $g\colon\tau^-(\Ker\alpha)\to T$, the morphism $f \circ g$ factors through $\alpha$. We denote by $\eta$ the conflation $\Ker\alpha\to X\xrightarrow\alpha Y$. We obtain that $\eta.(f \circ g)$ splits.
  Since $\Ker\alpha\in\CC_l$, there exists a natural isomorphism
  \[
    \phi\colon \Homcu(\tau^-(\Ker\alpha),-)
    \longrightarrow D\Extc(-,\Ker\alpha).
  \]
  Set
  $\gamma = \phi_{\tau^-(\Ker\alpha)}(\underline{\Id_{\tau^-(\Ker\alpha)}})$.
  By the naturality of $\phi$, we have
  \[
    \phi_T(\underline g)
    = D\Extc(g,\Ker\alpha)(\gamma)
    = \gamma\circ\Extc(g,\Ker\alpha).
  \]
  It follows that
  \[
    \phi_T(\underline g)(\eta.f)
    = \gamma((\eta.f).g)
    = \gamma(\eta.(f\circ g))
    = 0.
  \]
  We observe that $\phi_T(\underline g)$ runs over all maps in $D\Extc(T,\Ker\alpha)$, when $\underline g$ runs over all morphisms in $\Homcu(\tau^-(\Ker\alpha),T)$.
  It follows that $\eta.f$ splits. In other words, the morphism $f$ factors through $\alpha$. Then the result follows.
\end{proof}

Given an object $C$ and a morphism $f\colon X\to Y$, we denote by $\Im\Homc(C,f)$ the image of the induced map $\Homc(C,f)\colon\Homc(C,X)\to\Homc(C,Y)$. Generalizing
\cite[Corollary~XI.3.4]{AuslanderReitenSmalo1995Representation}
to an exact category $\CC$, we obtain the following existence theorem; compare
\cite[Theorem~XI.3.6]{AuslanderReitenSmalo1995Representation}
and
\cite[Theorem~4.4]{Ringel2013Auslander}.
Since $\CC$ does not have enough projectives, the treatment here is completely different.

\begin{theorem}\label{thm:exist}
  Let $C$ and $Y$ be two objects. Assume that $C\in\CC_r$ and $H$ is a left $\Gamma_C$-submodule of $\Homc(C,Y)$ satisfying $\CP(C,Y)\subseteq H$. Then there exists some deflation $\alpha\colon X \to Y$, which is right $C$-determined such that $\Ker\alpha\in\add(\tau C)$ and $H=\Im\Homc(C,\alpha)$.
\end{theorem}

\begin{proof}
  By the assumption we have $\tau C\in\CC_l$ and $\tau^-\tau C\simeq C$ in $\underline\CC$. Then there exists a natural isomorphism
  \[
    \phi\colon\Homcu(C,-)
    \longrightarrow D\Extc(-,\tau C).
  \]
  Set $\gamma=\phi_C(\underline{\Id_C})$. By the naturality of $\phi$, for each object $Z$ and each morphism $f\colon C\to Z$, we have
  \[
    \phi_Z(\underline f)
    = D\Extc(f,\tau C)(\gamma)
    = \gamma\circ\Extc(f,\tau C).
  \]
  Then for each $\mu\in\Extc(Z,\tau C)$, we have
  \[
    \phi_Z(\underline f)(\mu)
    = \gamma(\Extc(f,\tau C)(\mu))
    = \gamma(\mu.f).
  \]

  We set $\underline H=H/\CP(C,Y)$ and set
  \[
    \underline H^\perp
    = \set{\mu\in\Extc(Y,\tau C) \middle|
    \phi_Y(\underline h)(\mu) = 0
    \mbox{ for each }
    \underline h \in \underline H}.
  \]
  We observe that $\underline H^\perp$ is a right $\Gamma_C$-submodule of $\Extc(Y,\tau C)$. Here, for any $f\in\Gamma_C$ and $\mu\in\Extc(Y,\tau C)$, the action of $f$ on $\mu$ is given by $\tau(\underline{f}).\mu$.
  Then there exists finitely many $\eta_1,\eta_2,\dots,\eta_n$ in $\Extc(Y,\tau C)$ such that $\underline H^\perp = \sum_{i=1}^n \eta_i\Gamma_C$. Assume that $\eta_i$ is represented by the conflation $\tau C\to X_i\xrightarrow{\alpha_i} Y$ for each $i=1,2,\dots,n$. We have that $\alpha_i$ is right $\tau^-\tau C$-determined by Lemma~\ref{lem:ker.det}. Then $\alpha_i$ is right $C$-determined by Proposition~\ref{prop:det}. We observe that $\bigoplus_{i=1}^n \alpha_i$ is a deflation. By Lemma~\ref{lem:det.sum}, we have that $\bigoplus_{i=1}^n \alpha_i$ is right $C$-determined.

  Consider the following commutative diagram obtained by a pullback
  \[\xymatrix@+1pc{
    \bigoplus_{i=1}^n \tau C\ar@{-->}[r] &X\ar@{-->}[r]^\alpha\ar@{-->}[d]  &Y\\
    \bigoplus_{i=1}^n \tau C\ar[r] &\bigoplus_{i=1}^n X_i &\bigoplus_{i=1}^n Y.
    \ar[u];[]^-{\smatrix{\Id_Y\\[-3pt]\vdots\\[3pt] \Id_Y}}
    \ar[l];[]^{\bigoplus_{i=1}^n \alpha_i}
    \ar@{=}[llu];[ll]
  }\]
  We have that $\alpha$ is a deflation and $\Ker\alpha$ lies in $\add(\tau C)$. By Lemma~\ref{lem:det.PB}, the deflation $\alpha$ is right $C$-determined. By a direct verification, we have
  \[
    \Im \Homc(C,\alpha)
    = \bigcap_{i=1}^n \Im\Homc(C,\alpha_i).
  \]

  For each $i=1,2,\dots,n$, we set
  \[
    ^\perp(\eta_i \Gamma_C)
    =\set{\underline h \in \Homcu(C,Y) \middle|
    \phi_Y(\underline{h})(\mu) = 0
    \mbox{ for each }
    \mu \in \eta_i \Gamma_C}.
  \]
  We observe that $^\perp(\eta_i\Gamma_C)$ is a left $\Gamma_C$-submodule of $\Homcu(C,Y)$. We mention that $\CP(C,Y)\subseteq\Im\Homc(C,\alpha_i)$, since $\alpha_i$ is a deflation.
  We claim that
  \[
    ^\perp(\eta_i\Gamma_C)
    = \Im \Homc(C,\alpha_i)/\CP(C,Y).
  \]

  Let $h\colon C\to Y$ be a morphism in $\Im\Homc(C,\alpha_i)$. We obtain that $\eta_i.h$ splits. Then we have
  \[
    \phi_Y(\underline h)(\tau(\underline f).\eta_i)
    = \gamma((\tau(\underline f).\eta_i).h)
    = \gamma(\tau(\underline f).(\eta_i.h))
    = 0,
  \]
  for each $f\colon C\to C$. It follows that
  $\Im\Homc(C,\alpha_i)/\CP(C,Y)\subseteq {^\perp(\eta_i\Gamma_C)}$.

  On the other hand, let $h\colon C\to Y$ be a morphism such that $\underline h\in{^\perp(\eta_i\Gamma_C)}$. Then we have $\phi_Y(\underline h)(\tau(\underline f).\eta_i)=0$ for each $f\colon C\to C$. By the definition of $\tau$ in \cite[Section~3]{JiaoGeneralized}, we have the following commutative diagram
  \[\xymatrix{
    \Homcu(C,Y)\ar[r]^-{\phi_Y} &D\Extc(Y,\tau C)\\
    \Homcu(C,Y)\ar[r]^-{\phi_Y} &D\Extc(Y,\tau C).
    \ar[lu];[l]_{\Homcu(\underline f,Y)}
    \ar[u];[]^{D\Extc(Y,\tau(\underline f))}
  }\]
  By a diagram chasing, we have
  \[\begin{split}
    \phi_Y(\underline{h} \circ \underline{f})
    &= \phi_Y ( \Homcu(\underline{f}, Y) (\underline{h}) )\\
    &= (D\Extc(Y, \tau(\underline{f})) \circ \phi_Y) (\underline{h})\\
    &= \phi_Y(\underline{h}) \circ \Extc(Y, \tau(\underline{f})).\\
  \end{split}\]
  Then for each $\eta_i$, we have
  \[
    \phi_Y(\underline h\circ\underline f)(\eta_i)
    = \phi_Y(\underline h)(\tau(\underline f).\eta_i)
    = 0.
  \]
  It follows that
  \[
    \phi_C(\underline f)(\eta_i.h)
    = \gamma((\eta_i.h).f)
    = \gamma(\eta_i.(h\circ f))
    = \phi_Y(\underline{h\circ f})(\eta_i)
    = 0.
  \]
  We observe that $\phi_C(\underline f)$ runs over all maps in $D\Extc(C,\tau C)$, when $\underline f$ runs over all morphisms in $\underline\End_\CC(C)$. It follows that $\eta_i.h$ splits. In other words, the morphism $h$ factors through $\alpha_i$. We then obtain $h\in \Im\Homc(C,\alpha_i)$. It follows that
  ${^\perp(\eta_i\Gamma_C)}\subseteq\Im\Homc(C,\alpha_i)/\CP(C,Y)$.

  We observe that
  \[
    \underline H
    = {^\perp(\underline H^\perp)}
    = {^\perp(\sum_{i=1}^n\eta_i\Gamma_C)}
    = \bigcap_{i=1}^n {^\perp(\eta_i\Gamma_C)},
  \]
  where the first equality follows from the isomorphism $\phi_Y$. It follows that
  \[
    \underline H
    = \bigcap_{i=1}^n \Im \Homc(C,\alpha_i)/\CP(C,Y)
    = \Im \Homc(C,\alpha)/\CP(C,Y).
  \]
  Then the result follows since $\CP(C,Y)\subseteq H$.
\end{proof}

\section{A characterization for determined deflation}

Let $\CC$ be a Hom-finite Krull-Schmidt exact category. We give a characterization for a deflation being right $C$-determined for some object $C$.

The following lemma, which is due to
\cite[Proposition~XI.2.4 and Lemma~XI.2.1]{AuslanderReitenSmalo1995Representation},
shows that each right $C$-determined morphism has a minimal right determiner.

\begin{lemma}\label{lem:min.det}
  Let $f\colon X\to Y$ be a morphism, which is right $C$-determined for some object $C$.
  \begin{enumerate}
    \item\label{item:lem:C:1}
      Assume that $C'$ is an indecomposable object and $f'\colon C'\to Y$ almost factors through $f$. Then $C'$ is a direct summand of $C$.
    \item\label{item:lem:C:2}
      Assume that $\set{C_1,C_2,\dots,C_t}$ is a complete set of pairwise non-isomorphic indecomposable objects such that there exists some morphism $f_i\colon C_i\to Y$, which almost factors through $f$. Then $\bigoplus_{i=1}^t C_i$ is a minimal right determiner of $f$.
    \qed
  \end{enumerate}
\end{lemma}

Recall that a morphism $f\colon X\to Y$ is called \emph{right minimal}, if each $g\in\Endc(X)$ with $f\circ g=f$ is an automorphism. Dually, we call $f$ \emph{left minimal}, if each $g\in\Endc(Y)$ with $g\circ f=f$ is an automorphism.

Two morphisms $f\colon X\to Y$ and $f'\colon X'\to Y$ are called \emph{right equivalent} if $f$ factors through $f'$ and $f'$ factors through $f$. Assume that $f$ and $f'$ are right equivalent. Given an object $C$, we mention that $f$ is right $C$-determined if and only if so is $f'$. We observe that $\CC$ has split idempotents;
see \cite[Corollary~4.4]{Krause2015Krull}.
By \cite[Proposition~7.6]{Buhler2010Exact},
we have that $f$ is a deflation if and only if so is $f'$.

Given a morphism $f\colon X\to Y$ , we call a right minimal morphism $f'\colon X'\to Y$ the \emph{right minimal version} of $f$, if $f$ and $f'$ are right equivalent. Following
\cite[Section~2]{Ringel2012Morphisms},
we call $\Ker f'$ the \emph{intrinsic kernel} of $f$.

We mention that each morphism $f\colon X\to Y$ has a right minimal version; see
\cite[Theorem~1]{Bian2009Right}.
Indeed, by the projectivization in
\cite[Section~II.2]{AuslanderReitenSmalo1995Representation},
we have an equivalence $F \colon \add(X \oplus Y) \to \proj \Gamma_{X \oplus Y}$, where $\proj \Gamma_{X \oplus Y}$ is the category of finitely generated projective $\Gamma_{X \oplus Y}$-modules. By
\cite[Proposition~I.2.1]{AuslanderReitenSmalo1995Representation},
the morphism $F(f)$ has a right minimal version $g$ in $\proj \Gamma_{X \oplus Y}$. Then $F^{-1}(g)$ yields the required morphism, where $F^{-1}$ is a quasi-inverse of $F$.

Dually, two morphisms $f\colon X\to Y$ and $f'\colon X\to Y'$ are left equivalent if $f$ factors through $f'$ and $f'$ factors through $f$. We mention that there exists some left minimal morphism $g\colon X\to Z$ such that $f$ and $g$ are left equivalent. We call $g$ the \emph{left minimal version} of $f$, and call $\Cok g$ the \emph{intrinsic cokernel} of $f$.

\begin{lemma}\label{lem:exist.ass}
  Let $\alpha\colon X\to Y$ be a deflation and $C$ be an indecomposable object. Assume that there exists some morphism $f\colon C\to Y$ which almost factors through $\alpha$. Then there exists an almost split conflation $K\to E\to C$ such that $K$ is a direct summand of the intrinsic kernel of $\alpha$.
\end{lemma}

\begin{proof}
  We may assume that $\alpha$ is right minimal. Let $\beta\colon Z\to C$ be the pullback of $\alpha$ along $f$. Then $\beta$ is a deflation. By Proposition~\ref{prop:ras}, we have that $\beta$ is right almost split. Let $\gamma\colon E\to C$ be the right minimal version of $\beta$. We observe that $\gamma$ is a right almost split deflation and $\Ker\gamma$ is a direct summand of $\Ker\beta\simeq\Ker\alpha$.

  We claim that $\delta \colon \Ker\gamma \xrightarrow{\phi} E\xrightarrow{\gamma} C$ is an almost split conflation;
  see \cite[Proposition~II.4.4]{Auslander1978Functors}.
  Indeed, let $f \colon \Ker\gamma \to K'$ be a non-section. Assume that $f$ does not factor through $\phi$. Then the conflation $f.\delta$ does not split. In particular, the deflation $\gamma'$ does not split, and hence factors through $\gamma$. We then obtain the following commutative diagram
  \[\xymatrix{
    \llap{$\delta \colon$ } \Ker\gamma \ar[r]^\phi \ar[d]_f
    &E \ar[r]^\gamma \ar[d]_g
    &C \ar@{=}[d]
    \\
    \llap{$f.\delta \colon$\quad} K' \ar[r]^{\phi'} \ar@{-->}[d]_{f'}
    &E' \ar[r]^{\gamma'} \ar@{-->}[d]_{g'}
    &C \ar@{=}[d]
    \\
    \llap{$\delta \colon$ } \Ker\gamma \ar[r]^\phi
    &E \ar[r] ^\gamma
    &C.
  }\]
  Since $\gamma$ is right minimal, the morphism $g' \circ g$ is an isomorphism. We obtain that $f' \circ f$ is an isomorphism, which is a contradiction since $f$ is not a section. It follows that $f$ factors through $\phi$. We obtain that $\phi$ is left almost split, and then $\delta$ is an almost split conflation.
\end{proof}

\begin{corollary}\label{cor:min.det}
  A minimal right determiner of a deflation has no nonzero projective direct summands and lies in $\CC_r$.
\end{corollary}

\begin{proof}
  Let $C$ be a minimal right determiner of a deflation $\alpha\colon X\to Y$. It is sufficient to show that each indecomposable direct summand $C'$ of $C$ is non-projective and lies in $\CC_r$. By Lemma~\ref{lem:min.det}, there exists some morphism $f\colon C'\to Y$ which almost factors through $\alpha$. We observe that $C'$ is not projective, since $f$ does not factor through $\alpha$. By Lemma~\ref{lem:exist.ass}, there exists an almost split conflation ending at $C'$. Then the result follows from Lemma~\ref{lem:ass}(\ref{item:lem:X:1}).
\end{proof}

It is well known that in the category of finitely generated modules over an artin algebra, each morphism is right $C$-determined for some object $C$; see \cite[Corollary~XI.1.4]{AuslanderReitenSmalo1995Representation} and \cite[Theorem~1]{Ringel2012Morphisms}. Here, we can only give a characterization for a deflation being right $C$-determined for some object $C$.

\begin{theorem}\label{thm:det}
  A deflation $\alpha\colon X\to Y$ is right $C$-determined for some object $C$ if and only if the intrinsic kernel of $\alpha$ lies in $\CC_l$.
\end{theorem}

\begin{proof}
  For the sufficiency, let $\alpha'\colon X'\to Y$ be the right minimal version of $\alpha$. We observe that $\alpha'$ is still a deflation and $\Ker\alpha'\in\CC_l$. By Lemma~\ref{lem:ker.det}, we have that $\alpha'$ is right $\tau^-(\Ker\alpha')$-determined. It follows that $\alpha$ is right $\tau^-(\Ker\alpha')$-determined since $\alpha$ and $\alpha'$ are right equivalent.

  For the necessity, we may assume that $C$ is a minimal right determiner of $\alpha$. By Corollary~\ref{cor:min.det}, we have $C\in\CC_r$. We observe that $\Im\Homc(C,\alpha)$ is a $\Gamma_C$-submodule of $\Homc(C,Y)$. Since $\alpha$ is a deflation, we have $\CP(C,Y)\subseteq\Im\Homc(C,\alpha)$. By Theorem~\ref{thm:exist}, there exists some deflation $\alpha'\colon X'\to Y$, which is right $C$-determined such that $\Ker\alpha'\in\add(\tau C)$ and $\Im\Homc(C,\alpha)=\Im\Homc(C,\alpha')$. We observe that $\alpha\circ f$ factors through $\alpha'$ and $\alpha'\circ f'$ factors through $\alpha$ for any $f\colon C\to X$ and $f'\colon C\to X'$. Since $\alpha$ and $\alpha'$ are right $C$-determined, we have that $\alpha$ and $\alpha'$ factor through each other. It follows that $\alpha$ and $\alpha'$ are right equivalent. The intrinsic kernel of $\alpha$ is also the intrinsic kernel of $\alpha'$, which is a direct summand of $\Ker\alpha'$. Then the result follows since $\Ker \alpha' \in \add (\tau C)$ and $\tau C \in \CC_l$.
\end{proof}

\begin{example}\label{ex:det}
  Let $Q$ be the following infinite quiver
  \[
    \underset1\circ \longrightarrow
    \underset2\circ \longrightarrow
    \cdots \longrightarrow
    \underset n\circ \longrightarrow\cdots.
  \]
  We consider the representations of $Q$ over a field. For each $i\geq1$, we denote by $P_i$ the indecomposable projective representation and by $S_i$ the simple representation corresponding to $i$. Let $\CC$ be the category of finitely presented representations. It is well known that $\CC$ is a Hom-finite Krull-Schmidt abelian category.

  For each $i\geq1$, we consider the projective cover $f_i\colon P_i\to S_i$. We have $\Ker f_i\simeq P_{i+1}$. By \cite[Proposition~4.4]{JiaoGeneralized}, we have that $P_{i+1}\not\in\CC_l$. By Theorem~\ref{thm:det}, we have that $f_i$ is not right $C$-determined for any object $C$.

  By Lemma~\ref{lem:exist.ass}, for any indecomposable object $C$, any morphism $g\colon C\to S_i$ does not almost factor through $f_i$. We observe that $f_i$ is not right 0-determined. Then we have that the condition ``$f$ is right $C$-determined for some object $C$'' for Lemma~\ref{lem:min.det}(\ref{item:lem:C:2}) is necessary.
\end{example}

\section{More descriptions of objects in $\CC_r$ and $\CC_l$}

Let $\CC$ be a Hom-finite Krull-Schmidt exact category. We will give some characterizations for an object lying in $\CC_r$ or $\CC_l$ via morphisms determined by objects.

\begin{proposition}\label{prop:C_l}
  Let $K$ be an object without nonzero injective direct summands. Then $K$ lies in $\CC_l$ if and only if there exists some deflation $\alpha\colon X\to Y$, which is right $C$-determined for some object $C$ such that $K$ is the intrinsic kernel of $\alpha$.
\end{proposition}

\begin{proof}
  The sufficiency follows from Theorem~\ref{thm:det}. For the necessity, we assume $K\in\CC_l$. Decompose $K$ as the direct sum of indecomposable objects $K_1,K_2,\dots,K_n$. We have that each $K_i$ is non-injective. Then for each $i=1,2,\dots,n$, there exists a non-split conflation $K_i\to X_i \xrightarrow{\alpha_i} Y_i$. We observe that $\bigoplus_{i=1}^n \alpha_i$ is a deflation. By Lemma~\ref{lem:ker.det}, we have that $\bigoplus_{i=1}^n \alpha_i$ is right $\tau^-K$-determined. We observe that $\alpha_i$ is right minimal. Then $\bigoplus_{i=1}^n \alpha_i$ is also right minimal. It follows that $K$ is the intrinsic kernel of $\bigoplus_{i=1}^n \alpha_i$.
\end{proof}

The following lemma is the converse of Theorem~\ref{thm:exist}.

\begin{lemma}\label{lem:C_r}
  Let $C$ be an object. Assume that for each object $Y$ and each $\Gamma_C$-submodule $H$ of $\Homc(C,Y)$ satisfying $\CP(C,Y)\subseteq H$, there exists some deflation $\alpha\colon X\to Y$, which is right $C$-determined such that $H=\Im\Homc(C,\alpha)$. Then the object $C$ lies in $\CC_r$.
\end{lemma}

\begin{proof}
  It is sufficient to show that each non-projective indecomposable direct summand $C'$ of $C$ lies in $\CC_r$. Then each $f\in\CP(C,C')$ is not a retraction. We observe that $\radc(C,C')$ is formed by non-retractions. We obtain $\CP(C,C')\subseteq\radc(C,C')$. We observe that $\radc(C,C')$ is a $\Gamma_C$-submodule of $\Homc(C,C')$. By the assumption, there exists some deflation $\alpha\colon X\to C'$, which is right $C$-determined such that $\radc(C,C')=\Im\Homc(C,\alpha)$.

  We claim that $\Id_{C'}$ almost factors through $\alpha$. Indeed, the deflation $\alpha$ is not a retraction, since $\Im\Homc(C,\alpha) = \radc(C,C')$ is a proper submodule of $\Homc(C,C')$. It follows that $\Id_{C'}$ does not factor through $\alpha$. Let $f\colon T\to C'$ be a morphism in $\radc(T,C')$. For each $g\colon C\to T$, the morphism $f\circ g$ lies in $\radc(C,C')=\Im\Homc(C,\alpha)$. It follows that $f\circ g$ factors through $\alpha$. Since $\alpha$ is right $C$-determined, we have that $f$ factors through $\alpha$. It follows that $\Id_{C'}$ almost factors through $\alpha$. By Lemma~\ref{lem:exist.ass}, there exists an almost split conflation ending at $C'$. Then the result follows from Lemma~\ref{lem:ass}(\ref{item:lem:X:1}).
\end{proof}

Collecting the results obtained so far, we list some some characterizations for an object lying in $\CC_r$. We mention that the equivalence between Theorem~\ref{thm:C}(\ref{item:thm:C:5}) and Theorem~\ref{thm:C}(\ref{item:thm:C:6}) is somehow surprising.

\begin{theorem}\label{thm:C}
  Let $\CC$ be a Hom-finite Krull-Schmidt exact category and let $C\in\CC$. The following statements are equivalent.
  \begin{enumerate}
    \item\label{item:thm:C:1}
      The object $C$ lies in $\CC_r$.
    \item\label{item:thm:C:2}
      For each object $Y$ and each $\Gamma_C$-submodule $H$ of $\Homc(C,Y)$ satisfying $\CP(C,Y)\subseteq H$, there exists some deflation $\alpha\colon X\to Y$, which is right $C$-determined such that $H=\Im\Homc(C,\alpha)$.
    \item\label{item:thm:C:3}
      Each inflation $\alpha\colon X\to Y$ with $\Cok\alpha\in\add C$ is left $K$-determined for some object $K$.
  \end{enumerate}
  If moreover $C$ is non-projective indecomposable, they are equivalent to the following statements.
  \begin{enumerate}
    \setcounter{enumi}{3}
    \item\label{item:thm:C:4}
      There exists an inflation $\alpha\colon X\to Y$ whose intrinsic cokernel is $C$ such that $\alpha$ is left $K$-determined for some object $K$.
    \item\label{item:thm:C:5}
      There exists an almost split conflation ending at $C$.
    \item\label{item:thm:C:6}
      There exists a non-split deflation which is right $C$-determined.
    \item\label{item:thm:C:7}
      There exists a deflation $\alpha\colon X\to Y$ and a morphism $f\colon C\to Y$ such that $f$ almost factors through $\alpha$.
\end{enumerate}
\end{theorem}

\begin{proof}
  By Theorem~\ref{thm:exist} and Lemma~\ref{lem:C_r}, we have
  ``(\ref{item:thm:C:1}) $\Leftrightarrow$ (\ref{item:thm:C:2})''.

  By the dual of Lemma~\ref{lem:ker.det}, we have
  ``(\ref{item:thm:C:1}) $\Rightarrow$ (\ref{item:thm:C:3})''.
  For each indecomposable non-projective direct summand $C'$ of $C$, there exists some non-split inflation $\alpha \colon X \to Y$ such that $\Cok \alpha \simeq C'$. We observe that $\alpha$ is left minimal, since $\alpha$ is non-split and $C'$ is indecomposable. By the dual of Theorem~\ref{thm:det}, we have that $C' \in \CC_r$. Then
  ``(\ref{item:thm:C:3}) $\Rightarrow$ (\ref{item:thm:C:1})''
  follows.

  Now, we assume that $C$ is non-projective indecomposable.
  Then the dual of Proposition~\ref{prop:C_l} implies
  ``(\ref{item:thm:C:1}) $\Leftrightarrow$ (\ref{item:thm:C:4})'',
  and Lemma~\ref{lem:ass}(\ref{item:lem:X:1}) implies
  ``(\ref{item:thm:C:1}) $\Leftrightarrow$ (\ref{item:thm:C:5})''.

  It is well known that the right almost split deflation ending at $C$ is non-split and right $C$-determined; also see Lemma~\ref{lem:ker.det} and Proposition~\ref{prop:det}. Then we have
  ``(\ref{item:thm:C:5}) $\Rightarrow$ (\ref{item:thm:C:6})''.
  Let $\alpha$ be a non-split right $C$-determined deflation. We observe by Lemma~\ref{lem:min.det} that $C$ is a minimal right determiner of $\alpha$. By Lemma~\ref{lem:min.det}, we have
  ``(\ref{item:thm:C:6}) $\Rightarrow$ (\ref{item:thm:C:7})''.
  By Lemma~\ref{lem:exist.ass}, we have
  ``(\ref{item:thm:C:7}) $\Rightarrow$ (\ref{item:thm:C:5})''.
\end{proof}

\section{Exact categories having Auslander-Reiten duality}

Let $\CC$ be a Hom-finite Krull-Schmidt exact category. Following \cite[Definition~2.6]{Krause2013Morphisms} and \cite[Definition~3.1]{ChenLe2015note}, we introduce the following notion.

\begin{definition}\label{def:det}
  An exact category $\CC$ is said to \emph{have right stably determined deflations} if for each object $Y$ the following conditions hold.
  \begin{enumerate}
    \item\label{item:def:1}
      Each deflation ending at $Y$ is right $C$-determined for some object $C$.
    \item\label{item:def:2}
      For each object $C$ and each $\Gamma_C$-submodule $H$ of $\Homc(C,Y)$ such that $\CP(C,Y)\subseteq H$, there exists some deflation $\alpha\colon X\to Y$, which is right $C$-determined such that $H=\Im\Homc(C,\alpha)$.
    \qed
  \end{enumerate}
\end{definition}

Dually, an exact category $\CC$ is said to \emph{have left stably determined inflations} if the opposite category $\CC^{\op}$ has right stably determined deflations.

We give the following characterizations for an exact category having Auslander-Reiten duality in the sense of \cite{LenzingZuazua2004Auslander}.

\begin{theorem}\label{thm:AR}
  Let $\CC$ be a Hom-finite Krull-Schmidt exact category. The following statements are equivalent.
  \begin{enumerate}
    \item\label{item:thm:AR:1}
      $\CC$ has Auslander-Reiten duality.
    \item\label{item:thm:AR:2}
      $\CC$ has right stably determined deflations.
    \item\label{item:thm:AR:3}
      $\CC$ has left stably determined inflations.
  \end{enumerate}
\end{theorem}

\begin{proof}
  We observe that $\CC$ has Auslander-Reiten duality in the sense of \cite{LenzingZuazua2004Auslander} if and only if $\CC_l=\CC=\CC_r$.
  The dual of the equivalence between Theorem~\ref{thm:C}(\ref{item:thm:C:1}) and Theorem~\ref{thm:C}(\ref{item:thm:C:3}) implies that an object $K$ lies in $\CC_l$ if and only if each deflation whose kernel lies in $\add K$ is right $C$-determined for some $C$. Combining this with the equivalence between Theorem~\ref{thm:C}(\ref{item:thm:C:1}) and Theorem~\ref{thm:C}(\ref{item:thm:C:2}), we have
  ``(\ref{item:thm:AR:1}) $\Leftrightarrow$ (\ref{item:thm:AR:2})''.

  By duality, we have
  ``(\ref{item:thm:AR:1}) $\Leftrightarrow$ (\ref{item:thm:AR:3})''.
\end{proof}

\begin{remark}
  Compared with \cite[Theorem~3.4]{ChenLe2015note}, it is somewhat surprising that Theorem~\ref{thm:AR}(\ref{item:thm:AR:2}) and Theorem~\ref{thm:AR}(\ref{item:thm:AR:3}) are equivalent. It seems that Definition~\ref{def:det}(\ref{item:def:1}) and Definition~\ref{def:det}(\ref{item:def:2}) are more symmetric than the conditions (REC1) and (REC2) in \cite[Definition~3.1]{ChenLe2015note}.
\end{remark}

\section*{Acknowledgements}

We are very grateful to the referee for many helpful suggestions.
The first author thanks his advisor Professor~Xiao-Wu Chen for his guidance and encouragement.
We thank Dawei Shen for helpful comments.
The work is supported by National Natural Science Foundation of China (Nos. 11571329 and 11671174).

redpajama_set_name: RedPajamaArXiv


Jesus Sanchez "Master Printermaker Extradinaire"
Jesus is one of our favorite students here at Ventero. He is a student at Centennial School and participates in our R.A.I.N. program. We've been teaching him reductive block prints.
Reduction block prints get their name because you reduce or cut away your block for each new color. What you do is carve a little and print all your prints in the lightest/darkest color (depending on your paper color), then carve some more and print all your prints in the next lightest/darkest color, carve some more and print in a darker/lighter color yet, and so on. One thing to realize right at the beginning is that you can't go back and print more of the first color after you start carving for the second color! You will always be making a limited edition. If the second stage of your carving is the most beautiful stamp you've ever carved, and you go on to a third stage, tough luck! Print lots of copies right at the beginning -- more than you think you want, because you'll probably ruin a few in later stages.
Before I start my reduction print I need to already have a design in mind.
I'm going to print on black paper so I need to cut out everything that I want to remain black. Once I've done that I need to print my darkest color first (since I'm printing on black paper).
I've chosen to do these prints in three colors (red, orange and yellow). At this point I need to print my reds...here is an example of what that looks like.
Next, the final yellow layer, after cutting away the surface area that I wanted to remain orange.
Just wanted to remind everyone of a couple of important dates coming up. First one is March 21st, which is Open Mic Night. The second is March 28th, The Alamosa Art Walk. Ventero will have work displayed at Hospice del Valle. Hope to see you all at the different events.
Sign up and Support our Cause!


text:
iPhone Haute Links
We tested out this summer's newest fashion apps for the iPhone so you don't have to
By Rachel Raczka	• Published at 1:55 pm on August 19, 2009
We don't like to play favorites, but the iPhone may be one of our favorite toys of all time. And what thrills us even more is that the fashion industry is infiltrating the spectrum of apps with great promise. However, not all these apps are worth you time and effort, so we decided to take this summer's newest fashion related apps for a test drive.
PROS: The Chanel app is actually more than just visually appealing, they have "push" activated news tidbits and factoids for the camellia aficionado, including a reoccurring feature called, "Coco Chanel's Apartment," which explores the secrets behind her personal collection of homebound trinkets. Plus the classic logo looks just fantastic on our iPhone dashboard.
CONS: Honestly, we don't need the "Chanel Store Locator" feature and suspect they put it in there to fill space. Also, we were hoping for a Sims-esque Karl Lagerfeld avatar that yells at us in German when we've digested carbs.
StyleMixer, by GAP
PROS: We love these closet organization apps that make us feel like Cher from Clueless, but we hate paying for them. Fortunately for us, the great philanthropists over at the GAP have offered us a free version -- all we have to do is pilfer through their infiltration of the full GAP Fall collection into our own personal wardrobe, as well as full-screen adverts upon opening. Other than that, very easy to use and we like the design color scheme.
CONS: We actually don't own anything from the GAP and would prefer that they stop trying to use our GPS to tell us where the nearest location is. We already know. They're EVERYWHERE.
Fashion Finder, by SEVENTEEN MAGAZINE
PROS: We sort of like Rachel Bilson mini-me, Selena Gomez, who is featured in this month's version of the app, and on their current glossy cover.
CONS: Okay, we realize this app is made for the under-17-year-old, Seventeen-reading readership, but regardless, this app really has no substance whatsoever. They use the "NearbyNow Network" to locate the 61 products they featured in this month's issue. Unfortunately, nothing but items from Aeropostale and American Eagle Outfitters could be found in a 50 mile radius from Hartford. Go figure.
DK by Donna Karan New York
PROS: Donna Karan created content - a personal journal, travel blog, and it list of her own, called "Love Love Love." Plus, several built-in videos about style and the Donna Karan collection, starring -- take a guess -- Donna Karan, duh.
CONS: Does anyone really like Donna Karan this much?


My buddy Jim lives in Montreal & is a part of this great post-rock spaghetti western band called Golden Tombs. Anyway, they've just put out their debut album, but what I wanted to share with you all is their amazing video for the great cover they do of The Final Count Down. Watch, listen & enjoy!
This entry was posted in Music and tagged Golden Tombs.


text:
Liam Neeson cancels Unicef trip in protest at sacking of director
Hollywood legends Liam Neeson, Roger Moore and Vanessa Redgrave have withdrawn from Unicef work in protest at th...
CATHAL DERVAN
Irish Unicef leader Melanie Verwoerd sacked by email Laura Hutton/Photocall Ireland
Hollywood legends Liam Neeson, Roger Moore and Vanessa Redgrave have withdrawn from Unicef work in protest at the charity's decision to sack executive director Melanie Verwoerd.
Neeson has confirmed to the Irish Times that he pulled out of a trip to Mozambique this weekend after learning that Verwoerd, partner of the late broadcaster Gerry Ryan, has been dismissed.
Just last March Neeson became Unicef Ireland's first goodwill ambassador.
He told the paper that the sacking represented a 'profoundly damaging loss' to the charity.
The actor also appealed to Unicef to find another role for Verwoerd within the organization.
The statement, signed by Neeson and Redgrave, mother of his late wife Natasha, urged Unicef to re-employ Verwoerd so that 'she can continue her vocation and exceptional commitment to the children'.
Verwoerd has claimed she was sacked by email on July 15th because the board of Unicef felt publicity surrounding her relationship with the late Gerry Ryan was damaging the charity.
Irish UNICEF leader Melanie Verwoerd sacked by email
Liam Neeson is the first Irishperson to be appointed UNICEF "Goodwill Ambassador"
Golfer Rory McIlroy appointed as a UNICEF Ireland Ambassador
Liam Neeson and his new love Freya St Johnson papped on vacation
Unicef finally responded to the claims on Wednesday when it said the decision to dismiss Verwoerd was down to a disagreement as to her future at the organization.
Neeson and Redgrave told the Irish Times that they were 'profoundly disturbed' by the decision.
They also believe Verwoerd has been "hounded" by the Irish press but her commitment to Unicef had been 'unwavering'.
"The increase in donations to the charity last year was sure proof of her commitment," said the statement.
"It also proves that whatever various personal opinions, Melanie Verwoerd and the Irish Unicef committee have worked well to achieve this.
"Her commitment had inspired us and so many others to dedicate ourselves to the unique cause and history of Unicef.
"So firm is our trust in Melanie Verwoerd as an exceptional Unicef advocate and leader that we are compelled to say that we view the dismissal of Melanie as a profoundly damaging loss to Unicef."
Unicef Ireland said it respected Neeson's decision.