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It is well-known that the first homotopy group (fundamental group) of a manifold is countable. I would like to know this for higher homotopy groups of a manifold. i.e.

Question: Is it true that all homotopy groups of a manifold are countable?

I think one can do same process in the proof of countability of fundamental group for higher homotopy groups. i.e. using Lebesgue Number Lemma. See SM J. M. Lee. Am I right? If not any proof or reference for the proof.

C.F.G
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    I think the following two links give some information about this:

    https://mathoverflow.net/questions/201944/topological-n-manifolds-have-the-homotopy-type-of-n-dimensional-cw-complexes

    https://mathoverflow.net/questions/247899/a-cw-is-of-countable-type-iff-all-its-homotopy-groups-are-countable-reference

     – EBP Oct 01 '20 at 09:58
  • Thanks. The second link claims that my guess is wrong in general? – C.F.G Oct 01 '20 at 10:11
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    By combining both links, you can at least conclude that when the manifold is connected, then the homotopy groups are all countable, because every manifold has the homotopy type of a countable CW-complex, and connected countable CW-complexes have countable homotopy groups. Perhaps there is a counterexample to be found when considering non-connected manifolds. – EBP Oct 01 '20 at 10:40
  • But it seems to me that if the definition of Lee's book implies that the fundamental group is countable, then you have some countability axiom on the amount of connected components of your space perhaps. Then, it follows that all the higher homotopy groups also are countable. – EBP Oct 01 '20 at 10:49
  • @EBP: I think so, my question is about "manifold" and not all topological spaces. – C.F.G Oct 01 '20 at 10:52
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    The homotopy groups of a space only provide information about a single path-connected component, the component containing the base point. So it doesn't matter whether the manifold is connected. – John Palmieri Oct 01 '20 at 16:35

1 Answers1

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Every second-countable ANR has has the homotopy type of a countable CW complex. Thus every second-countable (i.e. metrisable) manifold, being an ANR, has the homotopy type of a countable CW complex.

Let $M$ be a second-countable manifold with a chosen basepoint. Note that $M$ is therefore separable and metrisable. Let $C_*(S^n,M)$ be the set of pointed maps $S^n\rightarrow M$. If $C_*(S^n,M)$ is given the uniform topology, then it becomes a separable metric space, and in particular is second-countable.

On the other hand, $C_*(S^n,M)$ in the compact-open topology is an ANR, and so homotopy equivalent to a CW complex. But since $S^n$ is compact, the compact-open and uniform topologies on $C_*(S^n,M)$ coincide. Therefore by the above $C_*(S^n,M)$ is homotopy equivalent to a countable CW complex. Thus it has countably many path-components, each of which is open. In particular $$\pi_0(C_*(S^n,M))\cong \pi_n(M)$$ is countable.

Tyrone
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