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Let $X$ be a finite CW complex and form the group $\mathcal{H}(X)$ of self-homeomorphisms $X\xrightarrow{\cong}X$, furnishing it with the compact-open topology. Under the assumptions on our space $\mathcal{H}(X)$ is a topological group. My very simple question is the following:

If $X$ is a finite CW complex, then does $\mathcal{H}(X)$ have CW homotopy type?

I couldn't find this question addressed in the literature. To my knowledge it is an open question as to whether $\mathcal{H}(X)$ is an ANR, even when $X$ is a (smooth) compact manifold (although it is in case that $\dim X=1,2$). Note however that I am asking for something much weaker.

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  • $\begingroup$ Just to be clear, are you asking about the homeomorphisms of the underlying topological space, or are you talking about homeomorphisms of the cell complex -- presumably this would be the subgroup of cellular homeomorphisms? $\endgroup$ – Ryan Budney Mar 17 at 18:14
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    $\begingroup$ @RyanBudney just the homeomorphisms of the underlying space. The question could be posed more generally, but finite CW complexes tend to be the things I end up caring about, and since they tend to have nice topological properties, it may make an answer more approachable. $\endgroup$ – Tyrone Mar 17 at 18:49
  • $\begingroup$ I guess there should be a way to "restrict" CW homotopy type of $Map(X,X)$ to $\mathcal{H}(X)$. $\endgroup$ – user43326 Mar 17 at 20:25
  • $\begingroup$ I'm guessing it's false without some regularity hypothesis on $X$. $\endgroup$ – Tom Goodwillie Mar 18 at 2:29

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