Let $f:G'\to G$ be a group homomorphism and let $A$ be a $G$-module.

We regard $A$ as a $G'$-module via restriction of scalars along $f$.

I am struggling to understand the following statement:

"$f$ induces group homomorphisms $H^q(G,A)\to H^q(G',A)$ for each $q\geq 0.$"

Concrete Interpretation

For each $q\geq 0$ there is a natural group homomorphism


determined by


These induce (as they compatible with the standard resolutions) group homomorphisms

$$H^q(G,A)\to H^q(G',A).$$


Doesn't this mean the induced map for $q=0$ is just the identity (edit: inclusion, not identity) map?

  • $\begingroup$ It is the inclusion, not the identity. $\endgroup$ – Mariano Suárez-Álvarez Oct 30 '16 at 18:22
  • $\begingroup$ Consider for example any nontrivla $G$-module $A$, let $G'$ be the trivial group and consider the inclusion $G'\to G$. $\endgroup$ – Mariano Suárez-Álvarez Oct 30 '16 at 18:23
  • $\begingroup$ Yes, of course, thank you! I think that's what I meant actually. But I'm still surprised, change of groups just leads to inclusion? $\endgroup$ – user350031 Oct 30 '16 at 18:47
  • $\begingroup$ That only works in degree $0$... Try to see what happens in positiv degree in the same situation in which $G'$ is trivial. $\endgroup$ – Mariano Suárez-Álvarez Oct 30 '16 at 18:50
  • $\begingroup$ Thank you - I will do! $\endgroup$ – user350031 Oct 30 '16 at 18:51

These induce (as they compatible with the standard resolutions) group homomorphisms $$H^q (G,A) \to H^q (G',A).$$

Well, actually, these morphisms have nothing to do with a particular resolution, they exist by some abstract nonsense. By definition, group cohomology is given by $$H^q (G,A) = (R^q (-)^G) (A)$$ —these are the right derived functors of the left exact functor of fixed points $$(-)^G\colon \mathbb{Z}[G]\textit{-Mod}\to \mathbb{Z}[G]\textit{-Mod}.$$ Now a group homomorphism $$f\colon G'\to G$$ gives us an exact forgetful functor $$f^\#\colon \mathbb{Z} [G]\textit{-Mod}\to \mathbb{Z} [G']\textit{-Mod}$$ —any $G$-module may be viewed as a $G'$-module via the action $g'\cdot x = f (g') \cdot x$. For the fixed points, it is easy to see that we get natural monomorphisms $$A^G \rightarrowtail (f^\# A)^{G'}.$$ Naturality means that for any morphism of $G$-modules $A\to B$ we have the corresponding commutative diagram. That is, we have a natural transformation $$H^0 (G, -) \cong (-)^G \Rightarrow (f^\# -)^{G'} \cong H^0 (G', f^\# (-)).$$ Now it follows from general properties of right derived functors (as they are universal $\delta$-functors, and $H^0 (G', f^\# (-))$ is a $\delta$-functor) that this uniquely extends to natural transformations in higher degrees $$H^q (G, -) \Rightarrow H^q (G', f^\# (-)).$$ These are not monomorphisms for $q > 0$, as the case $G' = \{ e \}$ suggests: the trivial group has trivial higher cohomology.

In general, if $T^q$ is a $\delta$-functor, then a natural transformation $R^0 F \Rightarrow T^0$ extends uniquely to natural transformations $R^q F\Rightarrow T^q$ that give you for any short exact sequence $0 \to K\to M \to N\to 0$ a commutative ladder with exact lines enter image description here


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