Could there be any homotopy group without "Lebesgue Number Lemma"?

Question

This is about a comment that I have made in my general topology class while I was proving the abovementioned lemma as a consequence of compactness!

As far as I know, essentially, there is only one proof of the fact that $$\pi_1S^1\simeq\mathbb{Z}$$ which depends on path-lifting and homotopy-lifting lemmata. Both of these lemmata, being the main tools in the theory of covering spaces, depend on the Lebesgue number to construct the liftings inductively. Consequently, it seems to me that the very basic machinery in the theory of covering spaces, or their counterparts when one tries to compute the fundamental group through action of discrete groups, were not available if we did not have the Lemma of Lebesgue. Therefore, noting that any higher homotopy group could be considered as the fundamental group of some iterated loop space, we could not have a reasonable computation of some higher homotopy groups without this lemma!

I wonder, if I am wrong, where the false claims of my argument live?

Answer 1

The Lebesgue Number Lemma is absolutely not needed to compute $\pi_1(S^1)$, or more generally to compute $\pi_n(S^n)$. Here's one way to do it that I've actually used while teaching several times.

The first step is to convince yourself that for smooth manifolds $M_1$ and $M_2$ equipped with basepoints, the set of homotopy classes of basepoint-preserving maps $M_1 \rightarrow M_2$ is the same as the set of smooth homotopy classes of basepoint-preserving smooth maps. This is easy (e.g. just convolve with a mollifier), and is worth knowing regardless.

You now have several options. The first is to do a baby version of the Pontrayagin-Thom argument to show that smooth maps $S^n \rightarrow S^n$ up to smooth homotopy are classified by their degrees. This gives you $\pi_n(S^n) = \mathbb{Z}$. Details can be found e.g. in Milnor's book on smooth manifolds. Showing that $\pi_k(S^n) = 0$ for $k < n$ is even easier since smooth maps $S^k \rightarrow S^n$ cannot be surjective (by, say, Sard's lemma).

If you just want $\pi_1(S^1) = \mathbb{Z}$, you can actually do it with covering spaces now without using the Lebesgue number argument. Instead, you prove smooth path-lifting for the universal cover $\mathbb{R} \rightarrow S^1$ by taking a smooth map $[0,1] \rightarrow S^1$ and integrating its velocity.

If you prefer to not use smooth methods, another option for computing $\pi_1(S^1)$ is to argue that since $S^1$ is a topological group, its fundamental group is abelian and hence is isomorphic to its first homology group (this doesn't use the Lebesgue number anywhere). You then compute $H_1(S^1)$ using whatever tools you want.

Answer 2

Lebesgue numbers are certainly not needed to compute $\pi_1(S^1)$ using lifting properties of covering spaces. I checked eleven books that compute $\pi_1(S^1)$ using the covering space ${\mathbb R}\to S^1$ and only four used Lebesgue numbers. The rest did not mention Lebesgue numbers and instead just used compactness. Of course compactness is used to prove the Lebesgue number lemma, but that does not mean that Lebesgue numbers must be used to prove lifting properties.

Similarly, Lebesgue numbers are not needed to prove the general lifting properties in covering space theory, nor are they needed for the proof of the van Kampen theorem. Compactness arguments suffice in both these situations.

As an alternative to using covering spaces to compute $\pi_1(S^1)$ one can use a more general version of the van Kampen theorem that applies when a space is decomposed as the union of two path-connected subspaces whose intersection is not assumed to be path-connected. In fact van Kampen's original 1932 paper includes such a generalization, although this seems to have been largely forgotten, as I have never seen this generalization mentioned in papers or books. (I am not talking about the use of groupoids to compute fundamental groups, which is quite a different story.) The more general van Kampen theorem deserves to be better known, and I will be including it in a revised version of my Algebraic Topology book that I am currently working on.