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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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).

Answer 5

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).

Answer 6

You may have a look at 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 Non standard Algebraic Topology