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?

  • 2
    $\begingroup$ 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 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.

  • $\begingroup$ 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 Jan 21 '13 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 Jan 23 '13 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 Jan 25 '13 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.