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• 1 Post By Deveno
• 1 Post By johnsomeone
• 1 Post By johnsomeone

Thread:

1. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

On pg 70 in Armstrong's Basic Topology book, the author writes:

" We introduce the disjoint union $X+Y$ of spaces $X,Y,$ and the function $j:X+Y \rightarrow X \cup Y$ which when restricted to either $X$ or $Y$ is just the inclusion in $X \cup Y$. This function is important for our purposes because:
a) it is continuous
b) the composition $(f \cup g)j: X+Y \rightarrow Z$ is continuous if and only if both $f$ and $g$ are continuous. "

The author does not mention anything about disjoint union before this. I also checked the index at the back of the book and did not find it. I found one definition of disjoint union on wiki. If I go by that definition then $X$ is not even a subset of $X+Y$ so I don't know what the author is trying to say when he writes '... the function $j:X+Y \rightarrow X \cup Y$ which when restricted to either $X$ or $Y$ ...'
Does anybody have a clue what is the meaning of disjoint union intended here and what this function $j$ might be?

2. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

there are different ways of defining the disjoint union. one way involves "tagging" X and Y that is:

X+Y = (Xx{1}) U (Yx{2}) (any singleton sets that are distinct could be used for the "tags").

if X,Y are already disjoint sets, then the disjoint union is simply XUY. but if X∩Y ≠ Ø, "tagging" the sets ensures that for an element a in X∩Y, we get two elements in X+Y:

namely (a,1) (from X) and (a,2) (from Y).

while, strictly speaking, X is not a subset of X+Y, we have the bijection:

x ↔ (x,1), so as SETS they are isomorphic.

the mapping j is the obvious one:

j(x) = x, if x is in X
j(y) = y, if y is in Y

perhaps an example will help:

let X = Y = R, the real number line. then R+R is two "separate" copies of the line (unlike say, RxR, which is a plane, or (Rx{0}) U ({0}xR), which is just the two coordinate axes). since a point in this space lies either on one line or the other, the image j(x) is just "whatever real number of the real line x belongs to is".

we might represent a point on the first line as (r,1), and a point on the second line as (s,2), if it happens that r = s, j maps both these points to r in RUR = R.

another, more abstract, way to characterize the disjoint union is as follows:

let X,Y be two topological spaces. then X+Y is a space with two continuous embeddings:

j1:X→X+Y
j2:Y→X+Y

such that if we have two continuous maps f:X→Z and g:Y→Z, there is a UNIQUE continuous map h:X+Y→Z with

hj1 = f
hj2 = g

this map h is often written f+g, each "component" keeps track of whether any z in the image of h came from X, or came from Y.

the point is that Xx{*}, where {*} is any singleton space, is naturally homemorphic to X:

f(x,*) = x and
g(x) = (x,*) are clearly both continuous (from a given topology on X, and the product topology on Xx{*}) (note that there is only ONE toplogy on {*} consisting of:

T = {Ø,*}. on a singleton space, ALL topologies (the indiscrete and the discrete, or whatever) are exactly the same, so in the product topology we have just two kinds of open sets:

UxØ = Ø, and Ux{*}, where U is open in X).

the idea is this:

suppose X = {Bob,Ted,Alice} and Y = {Bob,Carol,Ted}. let's form X' = {BOB,TED,ALICE} (which is really just the same set as X, we just capitalized everything).

then X+Y = {Bob,BOB,Ted,TED,ALICE,Carol}, and j is given by:

j(Bob) = Bob
j(BOB) = Bob
j(Ted) = Ted
j(TED) = Ted
j(ALICE) = Alice
j(Carol) = Carol

j "preserves" the elements of X or Y when they are unique, and "identifies" them if they lie in the intersection of X and Y.

3. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

Originally Posted by Deveno
there are different ways of defining the disjoint union. one way involves "tagging" X and Y that is:

X+Y = (Xx{1}) U (Yx{2}) (any singleton sets that are distinct could be used for the "tags").

if X,Y are already disjoint sets, then the disjoint union is simply XUY. but if X∩Y ≠ Ø, "tagging" the sets ensures that for an element a in X∩Y, we get two elements in X+Y:

namely (a,1) (from X) and (a,2) (from Y).

while, strictly speaking, X is not a subset of X+Y, we have the bijection:

x ↔ (x,1), so as SETS they are isomorphic.

the mapping j is the obvious one:

j(x) = x, if x is in X
j(y) = y, if y is in Y

perhaps an example will help:

let X = Y = R, the real number line. then R+R is two "separate" copies of the line (unlike say, RxR, which is a plane, or (Rx{0}) U ({0}xR), which is just the two coordinate axes). since a point in this space lies either on one line or the other, the image j(x) is just "whatever real number of the real line x belongs to is".

we might represent a point on the first line as (r,1), and a point on the second line as (s,2), if it happens that r = s, j maps both these points to r in RUR = R.

another, more abstract, way to characterize the disjoint union is as follows:

let X,Y be two topological spaces. then X+Y is a space with two continuous embeddings:

j1:X→X+Y
j2:Y→X+Y

such that if we have two continuous maps f:X→Z and g:Y→Z, there is a UNIQUE continuous map h:X+Y→Z with

hj1 = f
hj2 = g

this map h is often written f+g, each "component" keeps track of whether any z in the image of h came from X, or came from Y.

the point is that Xx{*}, where {*} is any singleton space, is naturally homemorphic to X:

f(x,*) = x and
g(x) = (x,*) are clearly both continuous (from a given topology on X, and the product topology on Xx{*}) (note that there is only ONE toplogy on {*} consisting of:

T = {Ø,*}. on a singleton space, ALL topologies (the indiscrete and the discrete, or whatever) are exactly the same, so in the product topology we have just two kinds of open sets:

UxØ = Ø, and Ux{*}, where U is open in X).

the idea is this:

suppose X = {Bob,Ted,Alice} and Y = {Bob,Carol,Ted}. let's form X' = {BOB,TED,ALICE} (which is really just the same set as X, we just capitalized everything).

then X+Y = {Bob,BOB,Ted,TED,ALICE,Carol}, and j is given by:

j(Bob) = Bob
j(BOB) = Bob
j(Ted) = Ted
j(TED) = Ted
j(ALICE) = Alice
j(Carol) = Carol

j "preserves" the elements of X or Y when they are unique, and "identifies" them if they lie in the intersection of X and Y.
Thank you so much Denevo for your invaluable help on this. I'd be stuck forever on that page if it weren't for you. It must have consumed a lot of your time to type so much. Thanks again.

4. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

I used Armstrong in the 1980s (the book is still around here somewhere - I think overall it's a wonderful undergrad intro to topology), and I still remember distinctly being troubled by this ("Where are they floating, these two disjoint sets?") and my profs reply. He gave me the vitually the exact answer Deveno just gave you. (Basically he said "If you're worried putting them somewhere, just think of them as Xx{0} and Yx{1} in Zx[0,1], where Z is some monster set containing both X and Y.")

5. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

Originally Posted by abhishekkgp
On pg 70 in Armstrong's Basic Topology book, the author writes:

" We introduce the disjoint union $X+Y$ of spaces $X,Y,$ and the function $j:X+Y \rightarrow X \cup Y$ which when restricted to either $X$ or $Y$ is just the inclusion in $X \cup Y$. This function is important for our purposes because:
a) it is continuous
b) the composition $(f \cup g)j: X+Y \rightarrow Z$ is continuous if and only if both $f$ and $g$ are continuous. "

The author does not mention anything about disjoint union before this. I also checked the index at the back of the book and did not find it. I found one definition of disjoint union on wiki. If I go by that definition then $X$ is not even a subset of $X+Y$ so I don't know what the author is trying to say when he writes '... the function $j:X+Y \rightarrow X \cup Y$ which when restricted to either $X$ or $Y$ ...'
Does anybody have a clue what is the meaning of disjoint union intended here and what this function $j$ might be?
I am stuck again. I am trying to show that (b) holds in the text quoted above.
In the above $X,Y$ were subsets of some topological space and $X,Y,X \cup Y$ all have the induced topologies. The topology on $X+Y$ is the disjoint union topology which I have taken from Disjoint union (topology) - Wikipedia, the free encyclopedia . $f:X \rightarrow Z$ and $g:Y \rightarrow Z$ are two functions.
Let $U$ be open in $Z$.

(only if part)
Assume $f, g$ are continuous.

$((f \cup g)j)^{-1}(U)$

$=j^{-1}({f^{-1}(U) \cup g^{-1}(U)})$

$=((f^{-1}(U) \cup g^{-1}(U)) \cap X) \times \{ 1 \} \, \cup \, ((f^{-1}(U) \cup g^{-1}(U)) \cap Y) \times \{ 2 \}$.

I need to show that $((f^{-1}(U) \cup g^{-1}(U)) \cap X)$ is open in $X$. (similarly for $Y$), which I am unable to do. Please help.

6. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

I also tried to do this step by step through the inverse functions, and keeping the various sets straight is a chore. I think it's easier to establish the main set relationship in one shot (my Claim#1 below), and then tackle the continuity claims of the proposition (my Claim#2 below).

I'll write * for composition, and j-1 for "j inverse", and same for all functions. I hope it's clear. Say X and Y are subspaces of W. Then X+Y is a subspace of of Wx[0,1] defined by X+Y = Xx{0} Union Yx{1}. Also, note that if X intersect Y is not empty (in W), then this f-union-g function is only well defined if f = g on that intersection.)

Have the following *sets* and *functions* (we're ignoring continuity for now): f:X->Z, g:Y->Z, j: X+Y -> X-Union-Y, and f-union-g:X-Union-Y -> Z, and the composition f-union-g * j: X+Y->Z.
Claim 1: Let U be any subSET of Z. Then (f-union-g * j)-1(U) = ( ( (f-1(U))x{0} ) union ( (g-1(U))x{1} ) ) in X+Y.
Proof: The standard way of showing that each is contained in the other (Let (s,t) be in (f-union-g * j)-1(U) in X+Y. Then... ). Doing it this way bypasses the headaches about what happens on X intersect Y inside X-Union-Y. It's a completely typical and straightforward this way.
Claim 2: (f-union-g * j) is continuous iff both f and g are.
Proof: It drops straight out from Claim 1.
=>) ASSUME (f-union-g * j) is continuous. Let U open in Z. Then (f-union-g * j)-1(U) is open in X+Y, so ( (f-1(U))x{0} ) union ( (g-1(U))x{1} ) ) is open in X+Y,
... so f-1(U) open in X and g-1(U) open in Y.
<=) ASSUME f and g are continuous. Let U open in Z. Then f-1(U) open in X and g-1(U) open in Y,
... so (f-union-g * j)-1(U) = ( (f-1(U))x{0} ) union ( (g-1(U))x{1} ) ) is open in X+Y.

7. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

Originally Posted by johnsomeone
Claim 1: Let U be any subSET of Z. Then (f-union-g * j)-1(U) = ( ( (f-1(U))x{0} ) union ( (g-1(U))x{1} ) ) in X+Y.
Thank you so much this did it.

8. Re: Armstrong's Basic Topology. Pg 70. definition not clear.

you eyes probably glazed over when i gave the "abstract" characterization of X+Y. mine did, when i first saw it. i was like: err, wut?

but it is actually helpful. recall that i said if we have ANY two (continuous) maps f:X→Z and g:Y→Z then we have a UNIQUE continuous map:

h:X+Y→Z with hj1 = f, and hj2 = g

(the ji are "embeddings" of X and Y in X+Y:

j1(x) = (x,1)

j2(y) = (y,2), ok?).

so it suffices to verify that:

[(fUg)j]j1 = f and
[(fUg)j]j2 = g

but this is clear since if x is only in X, then:

[(fUg)j]j1(x) = (fUg)j(x,1) = (fUg)(x) = f(x)UØ = f(x)

and if y is only in Y:

[(fUg)j]j2(y) = (fUg)j(y,2) = (fUg)(y) = ØUg(y) = g(y)

and if z is in X∩Y (where f and g agree):

[(fUg)j]j1(z) = (fUg)j(z,1) = (fUg)(z) = f(z)Ug(z) = f(z) (since f(z) = g(z) for all z in in X∩Y)

[(fUg)j]j1(z) = (fUg)j(z,2) = (fUg)(z) = f(z)Ug(z) = g(z) (see above).

then the construction of X+Y guarantees that (fUg)j is continuous (it's just the map f+g). on the other hand, if f+g is continuous, then clearly f and g must be, since they are compositions of continuous maps.

in fact, one of the least troublesome ways to DEFINE the disjoint union topology is to take it to be the FINEST topology such that the inclusions:

j1:X→X+Y
j2:Y→X+Y

are continuous. this automatically gives us "just the open sets we need" (and makes for a lot less "pre-image chasing").

***********

a little bit of overview (why you are doing this): suppose we have two squares (which, if we want to be completely formal about, we can regard as two copies of IxI, the unit square in R2).

as two separate squares, we have (IxI)+(IxI). now let's say we want to paste these two squares together, to get a 2x1 rectangle. so we take a homeomorph of the 2nd square, say:

[1,2]x[0,1] = JxI, which is still (for all intents and purposes), (IxI)+(IxI) ≅ (IxI)+(JxI).

but now we have some "overlap" (IxI)∩(JxI) = {1}x[0,1] (the common edge of our two rectangles). let's say we have an open set U in our rectangle S = (IxI)U(JxI), that crosses the (vertical) line x = 1. since we want the embeddings to be continuous, what we wind up with in the disjoint union topology, is "the part of U that lies in (IxI)" + "the part of U that lies in (JxI)". for example if U is a circle of radius 1/4 at the point (1,1/2), we have a half-circle in IxI and another half-circle in JxI, which are open sets in the subspace (relative) topologies of IxI and JxI.

roughly speaking, XUY is the quotient space (X+Y)/~ where ~ identifies the two parts of X∩Y in X+Y. in general, unions of sets may not behave well in topology, because there's no guarantee that the topology on X and the topology on Y are "compatible". considering X+Y takes care of this, the x's do their thing in the X part, and the y's do their thing in the Y part. of course, to finally get to XUY, we need to consider a quotient space (or identification space). these don't behave so well (quotient maps generally don't preserve connectedness/disconnectedness, for example), but at least we've "isolated" the problem to "the tricky part".
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