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Let $G$ be a (discrete) group and $H\le G$ a subgroup of finite index. Then there is a transfer map $$tr\colon\thinspace H^\ast(H;M)\to H^\ast(G;M) $$ in group cohomology, where $M$ is any $G$-module (see Brown's "Cohomology of groups", Chapter III).

I think this construction should be natural, in the following sense. Let $f\colon\thinspace G'\to G$ be a homomorphism such that $H':=f^{-1}(H)$ is of finite index in $G'$, and let $M$ be a $G$-module. Then the following diagram commutes, where the horizontal maps are transfers: $$ \begin{array}{ccc} H^\ast(H;M) & \to & H^\ast(G;M) \newline \downarrow f^\ast & & \downarrow f^\ast \newline H^\ast(H';f^\ast M) & \to & H^\ast(G';f^\ast M) \end{array} $$ Note that I do not want to assume that $(G':H')=(G:H)$ (however, I am willing to assume that $H\le G$ and $H'\le G'$ are normal, if necessary).

Does anyone know of a reference for this naturality?

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    $\begingroup$ Thinking in terms of covering spaces, I would guess that what you need to assume is that the map of sets $G'/H'\to G/H$ is a bijection. If $H$ is normal then you have the injectivity, but you do need the surjectivity, too. Think of the case when $f$ is the inclusion $H\to G$. $\endgroup$
    – Tom Goodwillie
    Commented Jan 21, 2013 at 14:25

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I don't believe this is true. Let $(G, H) = (\Sigma_3, C_3)$ and $f : C_3 \to \Sigma_3$. Then your square says that $$H^1(C_3;\mathbb{Z}/3) = \mathbb{Z}/3 \longrightarrow H^1(\Sigma_3;\mathbb{Z}/3) = 0\longrightarrow H^1(C_3;\mathbb{Z}/3) = \mathbb{Z}/3$$ is the identity, which is false.

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  • $\begingroup$ The problem is that the pullback of $EG'\to BG'$ via $Bf$ is only connected if $f$ is surjective. If $f$ is surjective, the naturality holds. $\endgroup$
    – Johannes Ebert
    Commented Jan 21, 2013 at 19:14
  • $\begingroup$ Very nice, thanks. I came close to writing "or counter-example" in the question (honest)! I'll try to think more about what is true, and may post here when I figure it out. $\endgroup$
    – Mark Grant
    Commented Jan 23, 2013 at 7:24
  • $\begingroup$ For what it's worth, it seems the best one can hope for is the double coset formula (Proposition III.9.5(iii) in Brown) which applies when $f\colon G'\to G$ is a monomorphism. $\endgroup$
    – Mark Grant
    Commented Jan 25, 2013 at 12:40

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