TITLE: Cohomological dimension of a homomorphism
QUESTION [11 upvotes]: 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.

REPLY [3 votes]: One more example refering to the second question: Let 
$$\Gamma = \left\lbrace \left. \begin{pmatrix} 
1 & \ast & \ast \newline   & 1 & \ast \newline  & & 1 
\end{pmatrix}  \right\vert\  \ast \in \mathbb{Z} \right\rbrace$$
be the group of integral upper triangular matrices with unit diagonal. $\Gamma$ fits into the non-split central extension $$0 \to \mathbb{Z} \to \Gamma \to \mathbb{Z}^2 \to 0$$ that corresponds to a generator $\epsilon  \in H^2(\mathbb{Z}^2;\mathbb{Z}) = \mathbb{Z}$. 
Claim 1: $cd(\Gamma) = 3$ 
Since $\mathbb{Z}$ resp. $\mathbb{Z}^2$ has cd $1$ resp. $2$, the LHS spectral sequence $E_2^{ij} = H^i(\mathbb{Z}^2;H^j(\mathbb{Z};M))$ shows $cd(\Gamma) \le 3$. Moreover, by positional reasons $E_\infty^{2,1}=E_2^{2,1}$ and in particular $E_\infty^{2,1} = \mathbb{Z}$ for $M = \mathbb{Z}$. Hence $cd(\Gamma) = 3$. 
Claim 2: The inflation map $\text{inf}: H^2(\mathbb{Z}^2;M) \to H^2(\Gamma;M)$ is zero. 
Since the image of inflation is just $E_\infty^{2,0} = \text{coker}(d_2^{0,1})$, it is sufficient to show that $d_2^{0,1}$ is surjective. Since the action of $\Gamma$ on $M$ is induced by the action of $\mathbb{Z}^2$, it follows that $\mathbb{Z}$ acts trivially on $M$. Futhermore, $\mathbb{Z}^2$ acts trivially on $H^\ast(\mathbb{Z};M)$ because $\mathbb{Z}$ is central. Therefore $E_2^{0,1} = Hom(\mathbb{Z},M)$.  
Let $\alpha \in Hom(\mathbb{Z},M)$. Then 
$$d_2^{0,1}: Hom(\mathbb{Z},M) \to H^2(\mathbb{Z}^2;M)$$
is given by $d_2(\alpha)= - \alpha^\ast(\epsilon)$ (well-known formula) where 
$$\alpha^\ast: H^2(\mathbb{Z}^2;\mathbb{Z}) \to H^2(\mathbb{Z}^2;M)$$ is induced by $\alpha$ on the coefficients. By using a projective resolution or by Poincare duality one easily sees that 
$$\alpha^\ast : \mathbb{Z} \to M/\lbrace gm-m \mid g \in \mathbb{Z}^2 \rbrace =: \bar{M}$$
is just $\alpha$ composed with the natural projection. Identifying $Hom(\mathbb{Z},M) = M$ now shows that $d_2^{0,1}: M \to \bar{M}$ is the natural projection and hence surjective.