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Let $G$ be a discrete group with universal principal bundle $EG\to BG$, and let $X$ and $Y$ be left $G$-spaces. An equivariant map $\overline{f}:X\to Y$ induces a fibre-preserving map $f:EG\times_G X\to EG\times_G Y$ between Borel constructions, as in the following diagram.

$\require{AMScd}$ \begin{CD} X @>\overline{f}>> Y\\ @V V V @VV V\\ EG\times_G X @>>f> EG\times_G Y \\ @V V V @VV V\\ BG @>=>> BG \end{CD}

I believe some sort of converse to be true. That is, if $f:EG\times_G X\to EG\times_G Y$ is a fibre-preserving map between Borel constructions, it induces a $G$-equivariant map $\overline{f}:X\to Y$. Here I am identifying $G=\pi_1(BG,\ast)$, which acts (up to homotopy) on the fibres.

Does anyone know of a reference to a precise statement along these lines?

Edit (in response to user51223's comment): I'm mainly interested in the existence question: Is it true that there exists an equivariant map $\overline{f}:X\to Y$ if and only if there exists a map $f:EG\times_G X\to EG\times_G Y$ over $BG$? (So $\overline{f}$ doesn't have to be the induced map on fibres.)

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    $\begingroup$ I think that the map from X to Y is in general only coherently homotopy equivariant. Have a look at Dror,Dwyer and Kan, Equivariant maps which are self homotopy equivalences, Propositions 2.2 and 2.3 (this paper is available on Bill Dwyer's webpage). $\endgroup$
    – Gustavo Granja
    Commented Apr 11, 2020 at 23:09
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    $\begingroup$ That $f$ always induces $\overline{f}$ is a fact right? And, your question is that if this map is $G$-equivariant? Or are you looking for any $G$-equivariant map that possible induced by $f$? $\endgroup$
    – user51223
    Commented Apr 12, 2020 at 12:26
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    $\begingroup$ @GustavoGranja: This is really helpful, thanks. After reading their 2.3 I realised that $f$ induces a $G$-map $EG\times X\to EG\times Y$ (which is obvious from covering space theory). Is this what is meant by "coherently homotopy equivariant"? $\endgroup$
    – Mark Grant
    Commented Apr 12, 2020 at 12:45
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    $\begingroup$ @user51223: I'm interested in both versions of the question (hence the "along these lines"), but only really need the weaker existence question. I've edited to make this clear. $\endgroup$
    – Mark Grant
    Commented Apr 12, 2020 at 12:47
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    $\begingroup$ @MarkGrant You are very welcome. By coherently equivariant homotopic I meant extending to a homotopy coherent map of diagrams indexed by the category $G$ with one object (and $G$ as the monoid of endomorphisms). I think this is equivalent to the map extending to $EG \times X \to Y$ as $EG\times X \to X$ is a cofibrant replacement for the diagram $X$. Note that since $EG$ is a terminal object in the homotopy category of free $G$-spaces giving a $G$-equivariant map from $EG\times X \to EG \times Y$ is really the same as giving a $G$-equivariant map form $EG \times X \to Y$. $\endgroup$
    – Gustavo Granja
    Commented Apr 12, 2020 at 14:05

2 Answers 2

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The answer to the question in the edit is no. Take $G=\mathbb Z$, $X$ to be a point and $Y=\mathbb R$ with the action $n\cdot x = x+n$. Then there are no equivariant maps from $X$ to $Y$ but there are many maps from $EG \times_G X = BG = S^1$ to the cylinder $EG \times_G Y$ over $BG$.

In this special case when $X$ is a point, an equivariant map $X \to Y$ is a fixed point, while a map $EG \times_G X \to EG \times_G Y$ is a homotopy fixed point. The fixed points include in the homotopy fixed points, but in general I don't think there is much one can say about the inclusion.

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Similar to Granja's answer: let $X$ be a point, and $Y = EG$, and let $f$ be induced by the diagonal on $EG$.

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