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Let $G$ be a finite group, and $X$ be a free $G$-space. Moreover, assume that $X$ has a homotopy type of a CW-complex. Does $X$ have $G$-homotopy type of a $G$-CW complex also?

Edit: My main motivation for this question is as follows. Milnor has shown that if $Y$ has a homotopy type of a CW-complex and X is a compact Hausdorff space, then the space of all continuous map $M(X, Y)$ from $X$ to $Y$, endowed with the compact-open topology has a homotopy type of a CW-complex as well. Now, let G be a finite group, and $X$ and $Y$ be a G-spaces. We can equip M(X, Y) to the $G$-action $(g, f)\to g\cdot f$ where $(g\cdot f)(x)=g\cdot f(g^{-1}x)$ for all $x\in X$. I am interested to know whether M(X, Y) has the homotopy type of a $G$-CW complex for nice $G$- spaces, i.e, $X$, and $Y$ are finite G-CW-complexes. Moreover, I am interested in particular in the case that this action is free. For example, this action is free on $M(\mathbb{S}^n, \mathbb{S}^k)$ where $n > k$, as by Borsuk-Ulam theorem there is no $\mathbb{Z}_2$-map form a higher dimensional sphere to a lower dimensional sphere (Here, we consider $\mathbb{S}^m$ as a free $\mathbb{Z}_2$-space with the antipodal action)

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  • $\begingroup$ Can you clarify what you mean with "free $G$-space"? $\endgroup$ – Denis Nardin Jun 21 at 8:55
  • $\begingroup$ @Denis Nardin, By free $G$-space I mean the action is free, i.e, g.x=x implies g=e. In the question is assumed that X has a homotopy type of a Cw-complex. Also, I ma interested in homotopy type not weak homotopy type. Thanks in advance for your time. $\endgroup$ – 123... Jun 21 at 8:59
  • $\begingroup$ The first example i would look at is $S^1$ with the $\mathbb{Z}$-action given by rotation with an irrational angle. $\endgroup$ – HenrikRüping Jun 21 at 21:26
  • $\begingroup$ I am a bit too drunk to actually look at it, but if I would look at examples, this is the first one. $\endgroup$ – HenrikRüping Jun 21 at 21:28
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There is a paper by Stefan Waner from 1980, I think it's called "Equivariant Classifying Spaces", in which he proves an equivariant version of Milnor's theorem. It might do what you want.

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    $\begingroup$ Thank you very much for your time. Your help is highly appreciated. Yeah, the full bibliography of the paper as follows: Waner, Stefan. "Equivariant homotopy theory and Milnor’s theorem." Transactions of the American Mathematical Society 258, no. 2 (1980): 351-368, and the answer of my question is in section 4 of this paper. $\endgroup$ – 123... Jun 22 at 12:04

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