## What properties make $[0,1]$ a good candidate for defining fundamental groups?

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

The title essentially says it all. Consider the category $\mathfrak{Top}_2$ of triples $(J,e_0,e_1)$ where $J$ is a topological space, and $e_i \in J$. There is an obvious generalization of the definition of homotopic maps. Suppose we have selected $(J,e_0,e_1)\in \mathfrak{Top}_2$. We could say that two continuous maps $f,g:X\to Y$ are "$J$-homotopic" if there is a continuous map $h:X\times J\to Y$ such that $h(x,e_0) = f(x)$ and $h(x,e_1) = g(x)$. We could then define $\pi_1 (X,x)$ to be the set of continuous maps $f:J\to X$ satisfying $f(e_0)=f(e_1)=x$, with $J$-homotopic maps identified. Here in order to define composition of paths in the naive way, we need to have picked some homeomorphism continuous map from $J$ to $({0}\times J\cup {1}\times J)/(({0},e_1) = (1,e_0))$, taking $e_0$ to $e_0\times 0$ and $e_1$ to $e_1\times 1$. I have two questions:

1. Can $([0,1],0,1)$ be characterized as an object in $\mathfrak{Top}_2$ in a purely categorical manner?

2. When is $\pi_1 (X,x)$ a group? For that matter, when is $\pi_1 (X,x)$ associative?

Essentially, the question comes down to: what properties of $[0,1]$ are needed in order to do homotopy theory?

-

1
To define multiplication of paths, you don't need a homeomorphism. You just need a map from $J$ to that quotient (taking $e_0$ and $e_1$ to the right things). – Tom Goodwillie Mar 25 2012 at 22:32

1
I think really the circle is the correct object for defining the fundamental group. What is important is that it is a cogroup object in the category of pointed spaces if you allow coassociativity up to homotopy. – Benjamin Steinberg Mar 26 2012 at 0:35

2
The question supposes that fundamental groups were defined before the unit interval... – Jon Beardsley Mar 26 2012 at 1:56

5
You don't need the unit interval at all to define the fundamental group, at least for "nice spaces, since then you can characterize the fundamental group(oid) of $X$ in terms of its covering spaces. Unfortunately, I don't know how to recognize when a space is "nice" (e.g., locally path connected and semi-locally simply connected) without using the unit interval. – Charles Rezk Mar 26 2012 at 15:48

1
I'd just like to put in a plea for Dedecker, Paul and Valderrama, Jerko, Graphes et cographes sur une cat\'egorie abstraite. Application `a l'homotopie, C. R. Acad. Sci. Paris S\'er. A-B, 262, 1966, A377--A380, for an early discussion of an "interval object". – Ronnie Brown May 3 2012 at 10:27
show 8 more comments

## 6 Answers

The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints.

Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the terminal object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$.

Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such.

The theorem is proved here, and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides.

-

1
Thanks! This is exactly the sort of thing I was looking for. – Daniel Miller Mar 26 2012 at 11:22

7
+1, it is the perfect answer to this question. Let me remark something which is even more striking (at least for me), but in fact more easy to prove: $([0,1],0,1)$ is the terminal bipointed set! Here, a bipointed set is just a set equipped with two distinct elements. Why striking? The category of bipointed sets is very, very basic and set-theoretic, but the unit interval $[0,1]$, or equivalently $\mathbb{R} \cup \{\pm \infty\}$ is the basic analytical object. But we can characterize this in terms of a simple universal property which only talks about sets ... – Martin Brandenburg Mar 26 2012 at 19:49

8
@Guillaume: I assume Martin didn't mean what he wrote. As you say, [0,1] (with its endpoints) isn't the terminal bipointed set; it's the terminal bipointed set J equipped with a map $J \to J \vee J$. (As before, "bipointed" must be taken to entail distinct basepoints.) Martin's observation is indeed very nice, and appears to have been first observed by Peter Freyd. – Tom Leinster Mar 26 2012 at 22:02

1
Unfortunaltey one cannot edit comments. Yes, I meant it is the terminal coalgebra with respect to the smash product endofunctor on bipointed sets. – Martin Brandenburg Mar 27 2012 at 8:26

2
More like a join than a smash. – Todd Trimble Mar 27 2012 at 23:45
show 6 more comments

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Rather than answering the question, I want to claim that it's the wrong question.

There is a way to define fundamental group(oid)s and other homotopical notions that makes no reference to the unit interval or any other bipointed space. This goes by names like "shape theory" or "Cech homotopy" or "toposic homotopy". The basic idea is that a "path" consists of a chain of overlapping open sets---a purely topological notion. There's a nice paper by John Kennison called "What is the fundamental group?" which approaches this idea in an elementary way (no toposes required).

This sort of "homotopy" can "see" things which homotopy defined in terms of the unit interval cannot. For instance, the Warsaw circle has trivial $\pi_1$ in the traditional sense, but Cech homotopy can see that it contains a loop. Same goes for the "long circle" (close up the long line into a circle).

One can then simply observe that if your space admits lots of maps into it from the unit interval (e.g. is locally path-simply-connected or whatever), then the fundamental group(oid) as defined in terms of open sets can equivalently be defined using maps out of the unit interval. So from this point of view, there is nothing fundamental about the unit interval; it's just that a lot of the spaces we care about do indeed contain lots of "paths", so that for them we can give a simpler definition of homotopy in terms of paths.

-

Very inspiring, thank you! – Martin Brandenburg Mar 26 2012 at 19:42

I only have a partial answer for 1. and a hopefully non-confusing answer to 2.

To start with, let us work with the fundamental groupoid, which is more, ahem, fundamental and better suited to generalisation. In particular, we can consider the set $\pi^J(X,a,b)$ of homotopy classes (rel endpoints) of maps $(J,0,1) \to (X,a,b)$, which is more natural in the setting you outline. This is, assuming it isn't empty, a torsor for the groups $\pi^J(X,a,a)$ and $\pi^J(X,b,b)$, so you're not really losing too much. But the more important structure is the whole groupoid.

The unit interval is at least weakly initial in the category of path-connected bipointed spaces and homotopy classes of maps (and we always have a torsor as above). If you don't assume path-connected, then the two-point set (with any of its topologies) can be allowed, but is completely useless in measuring homotopy. This is an important fact using $[0,1]$, and this can't be derived from formal homotopy theory. One could define $J$-connectedness for other bipointed spaces $J$, but the utility of such a definition is debatable unless you put in extra conditions, like making it a cylinder object.

The 'reason' we get a fundamental groupoid is that $[0,1]$ is an $A_\infty$ topological cogroupoid, namely a groupoid object in $Top^{op}$, up to homotopy, and then coherence of that up to homotopy, and so on, all the way up. Woah, I hear you say, that's a bit extreme. But it is true, and we can just focus on the first few layers.

First, we have a cocomposition $[0,1] \to [0,1]\sqcup_{1,0}[0,1]$ and a coidentity $[0,1] \to \ast$. Then instead of coassociativity, which would be the equality of the to obvious maps $$[0,1] \to [0,1]\sqcup_{1,0}[0,1]\sqcup_{1,0}[0,1],$$ we have a homotopy between these two maps. We also have a map $$[0,1] \sqcup_{1,0}[0] \to [0,1]$$ expressing the identity on the right, and a similar one on the left. Again, these aren't equal to the identity maps of $[0,1]$, but are homotopic to them. And again, we have coinverses up to homotopy. The choices of all these homotopies aren't important (although you can look up representatives in any book on algebraic topology), because the spaces of such homotopies are contractible.

When we want to involve another space and actually get $\Pi_1(X)$, what we do is hom this topological $A_\infty$-cogroupoid into the space $X$, and get an $A_\infty$-groupoid, and then we truncate it to a groupoid, by quotienting out by these homotopies that we have chosen (but remember the choices are unimportant). It is important that $[0,1]$ is path-connected, because this makes the $A_\infty$-cogroupoid contractible in certain technical ways which are important for generalisations to higher categories (most of the ideas in this answer come from Todd Trimble's work). For instance, in my thesis I defined a certain sort of fundamental bigroupoid which could be applied to topological stacks, and I relied heavily on the $A_\infty$-cogroupoid structure, because it was the only way I could prove I even had a fundamental bigroupoid (I confess I did have much more complicated interval objects than here).

-
Nice answer, but I don't feel that you are getting to the heart of my question. Showing that the unit interval behaves well "up to homotopy" is presupposing the fact that we have defined homotopy using the unit interval. If we, so to speak, we were not already using $[0,1]$ to do homotopy theory, then how could we identify the relevant categorical properties of $[0,1]$? By categorical here, I mean expressible in the category of topological spaces and continuous maps (not up to homotopy). – Daniel Miller Mar 26 2012 at 1:26
You need to be able to define 'contractible' somehow in order to make a lot of this machinery to work. In order to do that, you need some sort of model category (not necessarily a Quillen one) or some sort of category with a cylinder object. The unit interval can be defined as a terminal coalgebra of an endofunctor of the category of intervals, see the examples section at ncatlab.org/nlab/show/terminal+coalgebra, but this is not a homotopy-theoretic construction. Note also that it assumes linearly ordered, not merely bipointed as you have done. – David Roberts Mar 26 2012 at 1:40
I find the notation $[0, 1] \times_{0, 1} [0, 1]$ confusing, since it suggests a pullback; I expect what you mean is a pushout which glues the endpoint 1 of the first interval to the endpoint 0 of the second. On a different note, I think the linear orderedness might be a slight red herring, e.g., the unit interval in the Dedekind reals in any Grothendieck topos is the terminal coalgebra for the sort of double-gluing construction outlined in the first sentence of this comment (if done constructively correctly; see the Elephant), as a functor on bipointed objects. – Todd Trimble Mar 26 2012 at 2:05

5
It's not true that the two-point set is completely useless for measuring homotopy. "Homotopies" using the Sierpinski space (the non-discrete, non-codiscrete topology on the two-point space) detect exactly the specialization ordering. And there is a map from [0,1] to the Sierpinski space, so any specialization inequality induces a path in the traditional sense. For the "usual" sorts of spaces, the specialization ordering is boring, but for (e.g.) finite topological spaces, it contains all the homotopy information. – Mike Shulman Mar 26 2012 at 5:18

1
I'd like to draw attention to the paper: Dedecker, Paul and Valderrama, Jerko, Graphes et cographes sur une cat\'egorie abstraite. Application `a l'homotopie, C. R. Acad. Sci. Paris S\'er. A-B,262, 1966, A377--A380, as an early discussion of an interval object. – Ronnie Brown May 7 2012 at 10:08
show 1 more comment

The above answers explain what is needed for the definition of a fundamental group to make sense. Let me try to answer the question from a different angle and explain what properties of the interval are needed for this notion to be reasonably well-behaved.

I believe that the key property is that, intuitively speaking, "small pieces of $I$ look just like the whole thing". More precisely, the interval can be subdivided arbitrarily finely into smaller intervals, i.e. given any open cover $\mathcal{U}$ of $I$ there is a sequence $0 = t_0 < \ldots < t_m = 1$ such that for every $i$ we have $[t_{i - 1}, t_i] \subseteq U$ for some $U \in \mathcal{U}$. In some sense this is a stronger version of the observation that gluing two intervals yields a space that is again homeomorphic to the interval. (It is interesting that the universal property of the theorem mentioned by Tom Leinster already implies the "strong version" of the subdivision property even though it is stated purely in terms of the "weak version".) This is easily proven using the Lebesgue's Lemma and is the starting point of standard techniques for calculating with fundamental groups like the path lifting property for coverings or the Van Kampen Theorem. A similar property of cubes leads to similar techniques for higher homotopy groups.

I cannot think of any other space with a property of this kind we could use in place of $I$. However, it would be interesting to see if there is some analogy between this standard approach to the fundamental group and approaches to something like Čech fundamental group (which I am unfamiliar with).

-

You may have a look at http://mathoverflow.net/questions/80777/what-is-a-continuous-path for a very much related discussion.

My opinion, in two words, is that the main property of $[0,1]$ is that one can glue the intervals and obtain basically the same thing. I think that one can define the fundamental group as soon as one has a way to describe paths using an ordered set for which the concatenation of intervals is well-behaved. Let me give an explicit example: in A-homotopy theory of graphs, one uses the natural numbers as ordered set and defines the continuous paths to be finitely supported functions from $\mathbb N$ to the graph with the property that $f(i-1)$ is a neighbor of $f(i)$. Then defines homotopy equivalence and gets a group that has several good properties. If you are interested, you may find details in Chapter 2 of http://arxiv.org/abs/1111.0268

I never really went through the details but I am confident that one can do something similar also using the hyperreal numbers as ordered set and define for instance the fundamental group of *$\mathbb R^2$ and *$\mathbb R^3$. I think that there is some hope to prove that they are not homeomorphic. This fact does not seem easy to prove using classical notions, as you can see here http://mathoverflow.net/questions/86562/non-standard-algebraic-topology

-

You may be interested in a paper entitled "The Big Fundamental Group, Big Hawaiian Earrings, and Big Free Groups" by J. Cannon and G. Conner. In this paper, they talk about "big intervals" and use them to define a "big fundamental group". A "big interval" is a totally-ordered set (with the order topology) which is compact and connected (equivalently, a linear continuum with first and last point). It is true that [0,1] is terminal in the category of (non-degenerate) big intervals. (Which also means that it is initial since for big intervals I and J, a monomorphism $I\to J$ exists precisely when an epimorphism $J\to I$ exists).

-

For example the end-compactified long line? Though I don't think this has an interval (=preserving endpoints) map from the unit interval... – David Roberts Mar 7 at 7:12
No, if that's the kind of map you want, no there's no such map.  There is an order-preserving map (even that preserve endpoints), and there are plenty of continuous maps, but no continuous map that preserves endpoints, no. – Keith Penrod Mar 8 at 0:47
Also, I suppose I'm abusing the words "initial" and "terminal".  For every big interval there is a continuous (and order-preserving) map to [0,1] preserving endpoints, but it is not unique. – Keith Penrod Mar 8 at 2:25
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 89, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9303408861160278, "perplexity_flag": "head"}