# Homotopy fixed point on the fibre of a fibration over BG

I am currently reading the paper Homotopy Fixed Point Methods for Lie Groups and Finite Loop Spaces by Dwyer and Wilkerson (can be found here http://www.math.purdue.edu/~wilker/papers/analysis.pdf for example), and I'm having trouble proving a basic result in the appendix on homotopy fixed points, and any help would be much appreciated.

Assumption: All groups are discrete groups, and $$EG$$ and $$BG$$ are CW-complexes (for example coming from the usual nerve construction).

Lemma 10.4: Let $$W \rightarrow BG$$ be a (Serre) fibration with fibre $$F$$. Then there is a proxy action of $$G$$ on $$F$$ such that $$F^{hG}$$ is homotopy equivalent to the space of sections of $$W \rightarrow BG$$.

Here proxy action (as they defined it a couple of paragraphs before the statement of the lemma) just means that there is an ordinary homotopy equivalence $$F \simeq X$$ where $$X$$ is a $$G$$-space and $$F^{hG} := X^{hG}$$.

Following their proof, we're supposed to consider the pullback $$\tilde{W}$$ for the proxy action space $$\require{AMScd}$$ $$\begin{CD} \tilde{W} @>{h}>> W\\ @VVV @VV{p}V \\ EG @>{q}>> BG \end{CD}$$ and since $$EG$$ is contractible, we obtain that $$\tilde{W} \simeq F$$. Now we need to show that $$\tilde{W}$$ has the correct homotopy fixed point as required. I've tried two approaches that I can't seem to make work.

Method 1: Recall now that $$\tilde{W}^{hG} := Map^G(EG,\tilde{W})$$ the space of $$G$$-equivariant maps, and write $$\Gamma(W\xrightarrow{p} BG)$$ for the space of sections. So a point in $$\tilde{W}^{hG}$$ is just a commuting square $$\require{AMScd}$$ $$\begin{CD} EG @>{f}>> W\\ @V{a}VV @VV{p}V \\ EG @>{q}>> BG \end{CD}$$ where $$a$$ is a $$G$$-map and a point in $$\Gamma(W\xrightarrow{p} BG)$$ is just a commuting square $$\require{AMScd}$$ $$\begin{CD} EG @>{s}>> W\\ @V{id}VV @VV{p}V \\ EG @>{q}>> BG \end{CD}$$ So we can clearly get a map $$\Gamma(W\xrightarrow{p} BG) \rightarrow \tilde{W}^{hG}$$.

To get a map the other way, recall that the $$G$$-Whitehead theorem (for instance 6.4.2 of Benson's Representations and Cohomology V2) gives that any $$G$$-self-map between $$EG$$ is a $$G$$-homotopy equivalence. So we obtain a diagram $$\require{AMScd}$$ $$\begin{CD} EG @>{\bar{a}}>> EG @>{f}>> W\\ @V{id}VV@V{a}VV @VV{p}V \\ EG @>{id}>> EG @>{q}>> BG \end{CD}$$ where the left square commutes up to a $$G$$-homotopy, call it $$\gamma$$ say. Using this homotopy, that $$p$$ was a fibration, and looking at the diagram $$\require{AMScd}$$ $$\begin{CD} EG\times 0 @>{f}>> W\\ @VVV @VV{p}V \\ EG\times I @>>{\gamma}> BG \end{CD}$$ we can get a point in $$\Gamma(W\xrightarrow{p} BG)$$. My problem with this is that I don't know how to show that the two maps so defined give the required homotopy equivalence. In fact I have some doubts that this will work since I've made so many choices in the construction of the second map.

Method 2: This uses the usual identification of $$\tilde{W}^{hG}$$ with the space of sections of the canonical fibration $$\tilde{W}_{hG} \xrightarrow{\pi} BG$$ where $$\tilde{W}_{hG} := EG \times_G \tilde{W}$$ the homotopy orbit space/Borel construction. One can show that the natural map $$\tilde{W}_{hG} \xrightarrow{\bar{h}} W$$ is a homotopy equivalence. Then I sort of want to say that because of this homotopy equivalence we obtain $$\Gamma(\tilde{W}_{hG} \xrightarrow{\pi} BG) \simeq \Gamma(W \xrightarrow{p} BG)$$, but the problem with this is that the two fibrations are using different copies of $$EG$$.

## 1 Answer

In case anyone's interested, here's a solution due to a couple of friends.

Write $$\tilde{W}^{hG} = Map^G(EG,\tilde{W}) = Map^G(EG,EG)\times_{Map^G(EG,BG)}Map^G(EG,W)$$ and $$\Gamma(W\rightarrow BG) = *\times_{Map^G(EG,BG)}Map^G(EG,W)$$.

Claim: $$EG \xrightarrow{\simeq} *$$ induces $$Map^G(EG,EG) = (EG)^{hG} \xrightarrow{\simeq} (*)^{hG} = *$$, that is, that $$Map^G(EG,EG)$$ is contractible.

Given the claim, we then look at the diagram $$\require{AMScd}$$ $$\begin{CD} * @>>> Map^G(EG,BG) @<<< Map^G(EG,W)\\ @V{\simeq}VV @V{id}VV @V{id}VV\\ Map^G(EG,EG) @>>> Map^G(EG,BG) @<<< Map^G(EG,W)\\ \end{CD}$$ and use for example (the dual of) Hovey's cube lemma (Lemma 5.2.6 of https://web.math.rochester.edu/people/faculty/doug/otherpapers/hovey-model-cats.pdf), together with the fact that any space is fibrant and that the right horizontal maps are fibrations to get the required homotopy equivalence of pullbacks.

Proof of claim: The unique $$G$$-map $$EG \xrightarrow{\simeq} *$$ induces a $$G$$-map $$EG\times EG \xrightarrow{pr_1} EG$$ between free $$G$$-spaces, so by $$G$$-Whitehead, is a $$G$$-homotopy equivalence. So quotienting out by $$G$$ we obtain $$(EG)_{hG} \xrightarrow{\simeq} (*)_{hG} = BG$$ We know that in fact this gives a fibre homotopy equivalence $$\require{AMScd}$$ $$\begin{CD} (EG)_{hG} @>{\simeq}>> (*)_{hG} = BG\\ @VVV @VVV\\ BG @>{id}>> BG \\ \end{CD}$$ for example from page 52 of May's Concise Algebraic Topology https://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf. But then the space of sections of the two vertical fibrations are homotopy equivalent, and we know that homotopy fixed point spaces can be identified with such spaces of sections. $$\square$$

Comment: The result is quite interesting as it provides sort of a converse to the well-known fact that for any $$G$$-space $$X$$ the homotopy fixed points $$X^{hG}$$ can be identified with the space of sections to the fibration $$X_{hG} \rightarrow BG$$.