Let $X$ be a compact connected manifold and $\mathcal H(X)$ be the group of all homeomorphisms of $X$, equipped with the compact-open topology. Is the fundamental group of $\mathcal H(X)$ countable? Is it finitely generated?

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    $\begingroup$ Maybe it is locally contractible and every locally contractible Polish group has a countable fundamental group? $\endgroup$
    – YCor
    Commented Jul 9 at 18:31
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    $\begingroup$ It is locally contractible - this result is due to A. V. Černavskij (1969). $\endgroup$ Commented Jul 9 at 18:36
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    $\begingroup$ Sure, it is all about the identity component. $\endgroup$ Commented Jul 9 at 19:30
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    $\begingroup$ @KevinCasto yes, and this is an abuse of language to ignore basepoints. But topological groups have a canonical basepoint, so things are clear-cut. $\endgroup$
    – YCor
    Commented Jul 9 at 21:30
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    $\begingroup$ By Lemma 8.6 of arxiv.org/abs/2205.01755 it is countable. It need not be finitely-generated, e.g. by Example 3 of msp.org/pjm/1976/67-2/p09.xhtml. $\endgroup$
    – skupers
    Commented Jul 10 at 4:13

1 Answer 1


Let $X$ be a compact topological manifold. Since $\mathcal{H}(X)$ is a separable metric space and $S^n$ is compact, the mapping space $C(S^n,\mathcal{H}(X))$ in the compact-open topology is also a separable metric space.

Since $\mathcal{H}(X)$ is locally contractible topological group [Cer], it is locally equiconnected [Fox]. Since $S^n$ is compact Hausdorff, the space $C(S^n,\mathcal{H}(X))$ is also locally equiconnected [Mil, Le.3].

Now consider the evaluation fibration sequence $$ C_*(S^n,\mathcal{H}(X))\rightarrow C(S^n,\mathcal{H}(X))\xrightarrow{ev}\mathcal{H}(X). $$

Theorem (Heath [Hea]) If $p:E\rightarrow B$ is a Hurewicz fibration and $E,B$ are locally equiconnected metric spaces, then for any $x\in B$, the fibre $p^{-1}(x)$ is locally equiconnected. $\quad\blacksquare$

It follows that $C_*(S^n,\mathcal{H}(X))$ is locally equiconnected and hence locally contractible. Thus $C_*(S^n,\mathcal{H}(X))$ has open path components. Since it is a separable metric space, it has at most countably many path components. Thus we have:

Proposition If $X$ is a compact manifold, then for any $n\geq$ the group $\pi_n(\mathcal{H}(X))\cong\pi_0(C_*(S^n,\mathcal{H}(X)))$ is countable. $\quad\blacksquare$

The ingredients of the proof were that $\mathcal{H}(X)$ is separable metric and locally contractible. The first condition holds for any second-countable manifold (which is hemicompact), and Černavskii has results taking care of the second condition.

Proposition The statement of the previous proposition holds when $X$ is an open manifold which is the interior of a compact manifold. $\quad\blacksquare$


[Cer] A. Černavskii, Local contractibility of the group of homeomorphisms of a manifold, (Russian) Mat. Sb. (N.S.) 79 (121) (1969), 307-356.

[Fox] R. Fox, On fibre spaces. II, Bulletin of the American Mathematical Society. 49 (1943), 733–735.

[Hea] P. Heath, A Pullback Theorem for Locally-Equiconnected Spaces, Man. Math 55 (1986), 233-238.

[Mil] J. Milnor, On spaces having the homotopy type of a CW complex, Trans. Amer. Math. Soc. 90 (1959), 272-280.


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