One more example refering to the second question: Let $$\Gamma = \lbrace left\lbrace \left. \begin{pmatrix} 1 & \ast & \ast \newline & 1 & \ast \newline & & 1 \end{pmatrix} \mid right\vert\ \ast \in \mathbb{Z} \rbrace$$ 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}$$is just \alpha composed with the natural projection. Identifying Hom(\mathbb{Z},M) = M now shows that d_2^{0,1}: M \to \bar{M} is the natural projection and hence surjective. 1 One more example refering to the second question: Let$$\Gamma = \lbrace \begin{pmatrix} 1 & \ast & \ast \newline & 1 & \ast \newline & & 1 \end{pmatrix} \mid \ast \in \mathbb{Z} \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.