For G-spaces one has the property that fixed points of a join of two spaces is the join of the fixed points. I want to prove an analogue for homotopy fixed points. Is it true that $$ X^{hG} \ast Y^{hG} \simeq (X\ast Y)^{hG} $$ where $(-)^{hG}$ means homotopy fixed points.
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2$\begingroup$ I suspect the answer is no (the join is the homotopy pushout of the diagram X←X×Y→Y and the homotopy fixed points do not usually preserve homotopy colimits) but I do not have an obvious counterexample. An idea is to consider X and Y with the trivial G action: then the homotopy fixed points are just Map(BG,-) $\endgroup$– Denis NardinJul 13, 2017 at 13:41
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$\begingroup$ @DenisNardin Your observation also suggests that this may be cured by considering some appropriate $G$-equivariant pushout. Like, the universal $G$-equivariant homotopy between $G$-maps $X\times Y\to X\to Z$ and $X\times Y\to Y\to Z$ (does one also need higher coherences?). $\endgroup$– მამუკა ჯიბლაძეJul 15, 2017 at 13:52
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$\begingroup$ @მამუკაჯიბლაძე Genuine fixed points of course commute with the equivariant join, (they commute with all homotopy limits and colimits) but I am not sure how this could help. $\endgroup$– Denis NardinJul 15, 2017 at 15:06
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$\begingroup$ @DenisNardin Cannot it happen that some homotopy quotient of a free $G$-space acquires genuine fixed points? $\endgroup$– მამუკა ჯიბლაძეJul 15, 2017 at 16:34
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It is indeed not true. Let $G = \mathbb{Z}$ and let $X = Y$ be the set of two elements with the action of $\sigma$, the generator of $\mathbb{Z}$ switching the two points. For a space $Z$ with a $<\sigma> = \mathbb{Z}$ action, a point $z \in Z^{h\mathbb{Z}}$ is the same as a point $z_0 \in Z$ and a path between $z_0$ and $\sigma(z_0)$. Thus we have $X^{hG} = Y^{hG}= \emptyset$ and thus $X^{hG} \star Y^{hG} = \emptyset$. However, $X \star Y \sim S^1$. Since $S^1$ is path connected we have $(S^1)^{hG} \neq \emptyset$.
