Pullback morphism of a hyperplane inclusion is zero in the derived category

Question

Let $L \subset \mathbb{C}^n$ be a hyperplane and let $i:L \to \mathbb{C}^n$ be the inclusion. Since $i$ is proper, we have induced maps $i^*: H^k_c(\mathbb{C}^n) \to H^k_c(L)$, and these maps are zero for all $k$ simply because at every degree either one of the compactly supported cohomology groups is zero.

On the other hand, these maps in compactly supported cohomology can be seen to come from a map in the derived category. Indeed, the unit of the adjunction $(i^{-1}, Ri_*)$, together with the fact that $i_* = i_!$, gives a map $$ \alpha: R\Gamma_c \underline{\mathbb{Z}}_{\mathbb{C}^n} \to R\Gamma_c Ri_* i^{-1} \underline{\mathbb{Z}}_{\mathbb{C}^n} = R\Gamma_c \underline{\mathbb{Z}}_{L} $$ and the taking $H^k$ gives the homomorphisms from before, i.e. $H^k(\alpha) = i^*$. My question is whether $\alpha = 0$ (seen as a map in the derived category $D^b(\textbf{Ab})$). Of course, this is stronger than what we said so far, which is $H^k(\alpha) = 0$ for all $k$, see this question.

Edit: if instead of $R\Gamma_c$ we take $R\Gamma$ then the map is nonzero (because it induces an isomorphism after taking $H^0$), so this is fundamentally a question about cohomology with compact support.

Answer 1

You are asking whether a specific element of ${\rm Hom}_{D^b({\rm Ab})}(\mathbb Z[-2n], \mathbb Z[-2n-2])$ is zero. But this group is ${\rm Ext}^2_{\rm Ab}(\mathbb Z, \mathbb Z) = 0$. (Because $\mathbb Z$ is projective, or because the cohomological dimension of abelian groups is $1$).

There are analogous settings where the map is nonzero in the derived category. You can consider the inclusion of a real hyperplane and consider coefficients in $\mathbb F_2 [ \Sigma_2]$ where the symmetric group $\Sigma_2$ acts by multiplying by signs. Or you could consider the inclusion of a complex hyperplane and take $U(1)$ (or equivalently $\mathbb C^*$)-equivariant coefficients