# Homotopy type of G-CW-structure

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)

• Can you clarify what you mean with "free $G$-space"? – Denis Nardin Jun 21 at 8:55
• @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. – 123... Jun 21 at 8:59
• The first example i would look at is $S^1$ with the $\mathbb{Z}$-action given by rotation with an irrational angle. – HenrikRüping Jun 21 at 21:26
• I am a bit too drunk to actually look at it, but if I would look at examples, this is the first one. – HenrikRüping Jun 21 at 21:28