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I have the following problem. Let $\Gamma_{G_1\times G_2}$ be a full subcategory of the orbit category $\mathcal{O}_{G_1\times G_2}$ consisting of graph subgroups of $G_1\times G_2$. Further, let $N$ be a dual (i.e., covariant) coefficient system over $G_1$ and define the dual coefficient system $FN$ over $G_1\times G_2$ to be: $ FN(\Delta)=N(\pi\Delta) $ if $\Delta$ is a graph subgroup and $0$ otherwise. Here $\pi\colon G_1\times G_2\to G_1$ is a projection on the first factor.

Let $X$ be a $G_1\times G_2$-space which is $1\times G_2$-free. I would like to show that $$ H^{G_1\times G_2}_\ast(X,FN)\cong H^{G_1}_\ast(\frac{X}{1\times G_2},N). $$

It's easy to see that this is true on $G_1\times G_2$-orbits (at least under graph subgroups) and I am pretty sure that there is some classical argument which can be used to prove this for general $G_1\times G_2$-spaces which are $1\times G_2$-free, but I fail to see it. It seems to be fairly similar to the argument that the ordinary homology (non-equivariant) with given coefficients is unique and I remember that there is some spectral sequence argument for this proof - so also a reference for this fact would be very useful.

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TL; DR: Suppose you have a functor between small categories $\mathcal C_0\to \mathcal C$. Let $\mathcal D$ be a locally small category such as Top or Ch. Then there is an adjunction of functor categories $$ L:[\mathcal C_0, \mathcal D]\leftrightarrows [\mathcal C, \mathcal D]:R$$ Where the right adjoint $R$ is the restriction and $L$ is the left Kan extension. There is a similar adjunction between categories of contravariant functors. Now suppose we have functors $F\colon \mathcal C_0\to \mathcal D$ and $G\colon \mathcal C^{\operatorname{op}}\to \mathcal D$. Then there is an isomorphism of coends: $$ F\otimes_{\mathcal C_0} RG\cong LF \otimes_{\mathcal C} G.$$

The isomorphism you ask about is an example of this adjunction.

------ Now with more details -------

By Elmendorf-type theorem, you can identify the category of $G_1\times G_2$-spaces that are $G_2$-free with the category of functors $[\Gamma_{G_1\times G_2}^{\operatorname{op}}, \mbox{Top}]$, where Top is the category of pointed spaces. Similarly, you can identify coefficient systems of the kind you consider with functors $[\Gamma_{G_1\times G_2}, \mbox{Ch}]$, where Ch denotes the category of chain complexes. More precisely, coefficient systems correspond to functors that take values in chain complexes concentrated in degree zero.

The group homomorphism $G_1\times G_2\to G_1$ induces a quotient by $G_2$ functor $\Gamma_{G_1\times G_2}\to \mathcal O_{G_1}$.

This functor induces an adjunction of functor categories $$[\Gamma_{G_1\times G_2}^{\operatorname{op}}, \mbox{Top}] \leftrightarrows [\mathcal O_{G_1}^{\operatorname{op}}, \mbox{Top}]$$ where the right adjoint is the pullback. The left adjoint, a.k.a the left Kan extension, can be identified, once again, with the quotient by $G_2$ functor $X\mapsto \frac{X}{1\times G_2}$. Here I have identified the functor categories with spaces with action of $G_1\times G_2$ and of $G_1$ respectively.

Similarly there is a pair of adjoint functors between categories of coefficient systems

$$[\Gamma_{G_1\times G_2}, \mbox{Ch}] \leftrightarrows [\mathcal O_{G_1}, \mbox{Ch}].$$ Here, again, the right adjoint is the pullback, and it is equivalent to the functor $F$ that you describe.

Finally, the Bredon homology groups $H_*^{G_1\times G_2}(X, FN)$ can be identified with the homology groups of the coend $$C_*\left(X^H\right)\otimes_{H\in \Gamma_{G_1\times G_2}} FN$$ and similarly the Bredon homology groups $H_*^{G_1}\left(\frac{X}{1\times G_2}, N\right)$ can be identified with the homology groups of the following coend $$C_*\left(\frac{X}{1\times G_2}^H\right)\otimes_{H\in \mathcal O_{G_1}} N.$$ The equivalence of the two coends follows from the adjunction.

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  • $\begingroup$ Thank you for the answer! However, I need one explanation here - how o you prove the formula for coends at the beginning? (Reference would also do) $\endgroup$ Commented Jun 24, 2021 at 10:13
  • $\begingroup$ Ok, I posted it as a separate question: mathoverflow.net/questions/396137/isomorphism-of-coends $\endgroup$ Commented Jun 25, 2021 at 7:53

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