# Naturality of the transfer in group cohomology

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?

• 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 $H\to G$. – Tom Goodwillie Jan 21 '13 at 14:25

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.
• 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. – Johannes Ebert Jan 21 '13 at 19:14
• 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. – Mark Grant Jan 25 '13 at 12:40