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$\DeclareMathOperator\Map{Map}\newcommand{\B}{\mathrm{B}}\newcommand{\h}{\mathrm{h}}$Let $f:G\to H$ be a morphism of topological groups and let

$$H^{\h G}:=\Map_G(\mathrm{E}G, H)$$

be the homotopy centralizer. Here $H$ is considered a $G$-space via the map $f$, that is, $g\cdot h=f(g)hf(g^{-1})$, and the object $H^{\h G}$ retains the structure of a topological group by pointwise multiplication.

Is it generally true that the classifying space $\B(H^{\h G})$ of $H^{\h G}$ satisfies

$$\B(H^{\h G})\simeq \Map(\B G, \B H)_{\B f}?$$

Here the right-hand side is the connected component corresponding to $f$ in the (unbased) mapping space between the classifying spaces of $G$ and $H$.

I have seen special cases of this statement in the literature, for example [Dwyer–Zabrodsky, "Maps between classifying spaces", Theorem 1.1]. But is the general statement above also true?

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1 Answer 1

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$\newcommand{\S}{\mathcal S}$ The answer is yes, always. Let's compute $\Omega(map(BG,BH),Bf)$.

The idea is to realize the map $* \xrightarrow{Bf} map(BG,BH)$ as a map of the form $\Gamma(BG,-)$ where $\Gamma$ means sections.

In more detail, consider the map $BG\xrightarrow{(id,Bf)} BG \times BH$ in $\S_{/BG}$.

Taking sections $\S_{/BG}\to \S$ gives exactly the map $* \to map(BG,BH)$ picking out $Bf$.

In particular, because sections commutes with (homotopy) pullbacks, you can compute the loop space before taking sections, so that in the end you're looking for sections of $BG\times_{BG\times BH} BG$, as a space over $BG$.

I claim that this is $H_{hG}$, i.e. the total space of the fibration classifying the conjugation action of $G$ on $H$. Indeed, by definition it is pulled back from $BG\to BG \times BH$, which itself is pulled back from the diagonal map $BH\to BH\times BH$, and the diagonal map certainly classifies the conjugation action of $H$ on itself.

So in total, it classifies the restriction along $G\to H$ of the conjugation action of $H$ on itself.

It follows at once that the space of sections is the homotopy limit, i.e. $H^{hG}$.

So this provides an equivalence of spaces $H^{hG}\simeq \Omega(map(BG,BH),f)$, and in particular a way of equipping $H^{hG}$ with a group structure, for which the statement on classifying spaces is obvious. If you have an a priori group structure on $H^{hG}$, and you want to check that this equivalence is compatible with it, it would take a bit more work.

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  • $\begingroup$ That is a nice answer, thanks a lot! For the second part of your argument I also found this nlab page helpful: ncatlab.org/nlab/show/free+loop+space+of+classifying+space It is shown there that $BH\times_{BH\times BH} BH \simeq H_{hH}$ (for the adjoint action), and pulling this back to $BG$ yields $BG\times_{BH\times BH} BH \simeq BG\times_{BG\times BH} BG$. $\endgroup$
    – Thomas
    Commented Feb 8 at 11:26

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