Let $G$ and $\Gamma$ be discrete groups, and let $\phi\colon\thinspace G\to \Gamma$ be a homomorphism.


Define its *cohomological dimension* $\operatorname{cd}\phi$ to be the least integer $d$ such that $\phi^\ast\colon H^i(\Gamma;M)\to H^i(G;M)$ is the zero homomorphism for all $i>d$ and all $\Gamma$-modules $M$ (where $M$ is regarded as a $G$-module via $\phi$).


Given that cohomological dimension of groups is such a well-studied invariant, I would have expected to find references to this relative notion in the literature. Alas, I cannot.



> 
> Are there any references considering cohomological dimension of homomorphisms?
> 
> 
> 


and more specifically



> 
> Does anyone know an example of a *surjective* homomorphism $\phi$ as above for which
> $$\operatorname{cd} \phi < \min\lbrace \operatorname{cd}G, \operatorname{cd} \Gamma \rbrace?$$
> 
> 
> 


**EDIT:** Thanks to Tom and Ralph's answers, I have been able to prove the following precise statement:


Let
$$ 1\to A \to G \stackrel{\phi}{\to} \Gamma \to 1$$
be a central extension, where $H\_\ast(A)$ is free and of finite type, and $\Gamma$ is a duality group with $\operatorname{cd}\Gamma = n$. Then $\operatorname{cd}\phi = n$.


**Proof.** We will show that $0\neq \phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma)\to H^n(G;\mathbb{Z}\Gamma)$. This follows from the Lyndon-Hochschild-Serre spectral sequence. Since the action of $\Gamma$ on $A$ is trivial, and $\mathbb{Z}\Gamma$ is a trivial $A$-module, the $E\_2$ term has
$$H^p(\Gamma;H^q(A;\mathbb{Z}\Gamma))\cong H^p(\Gamma;H^q(A)\otimes\mathbb{Z}\Gamma)$$
in the $(p,q)$-position. Since $\Gamma$ is a duality group, this is zero for $p\neq n$. Hence there are no non-trivial differentials, and the edge homomorphism
$$\phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma) \to H^n(G;\mathbb{Z}\Gamma)$$
is an isomorphism. $\Box$


Tom's answer shows that either centrality or finite type is necessary in the above statement. I haven't accepted it yet because I'm hoping someone will give an example with $\operatorname{cd} G <\infty$.


**Update:** Dranishnikov and Kuanyshov have investigated cohomological dimensions of homomorphisms more fully in a recent preprint <https://arxiv.org/abs/2203.03734>, in particular in relation to the Lusternik--Schnirelmann category of the classifying map $B\phi: BG\to B\Gamma$. They provide simpler examples of surjective $\phi$ with $\operatorname{cd}\phi<\min\{\operatorname{cd}G,\operatorname{cd}\Gamma\}$, including examples where $G$ and $\Gamma$ are geometrically finite. They also show that this phenomenon can't occur when $G$ and $\Gamma$ are torsion-free nilpotent groups.



Cohomological dimension of a homomorphism
How about this? Let $\Gamma$ be free abelian of rank $2$. Poincare duality in the torus identifies $H^2(\Gamma;M)$ with $H\_0(\Gamma;M)$, so that in particular there is a natural surjection $M\to H^2(\Gamma;M)$. 


Let $F$ be the particular $\Gamma$-module $\mathbb Z\Gamma$, free module of rank one for the group ring. A generator of $H^2(\Gamma;F)=H\_0(\Gamma;F)=\mathbb Z$ determines an extension of $\Gamma$ by $F$. Call it $G$. 


$\Gamma$ has cohomological dimension $2$. $G$ has cohomological dimension at least $2$, doesn't it? (EDIT: Yes, just because it has a subgroup $F$ with infinite cd.) But for every $\Gamma$-module the map $H^2(\Gamma;M)\to H^2(G;M)$ is zero, because using that natural surjection $M\to H^2(\Gamma;M)$ it's enough to prove this in the case $M=F$, where it is clear.


EDIT: To spell out this last step, there is a surjective map $M\to H\_0(\Gamma;M)$ (for every group $\Gamma$), natural in $M$. For this particular group, there is also a natural isomorphism $H\_0(\Gamma;M)\to H^2(\Gamma;M)$. Following this by your natural map $H^2(\Gamma;M)\to H^2(G;M)$, we get a map $M\to H^2(G;M)$, natural in $M$, and we just have to show that it takes every element $x\in M$ to $0$. But there is a map $F\to M$ taking a generator to $x$, so by naturality it is enough to see that you get the zero map when $M$ is $F$.


