# Does local cohomology commute with pullback?

Let $$Y$$ be a topological spaces and $$Z \subset Y$$ be locally closed, i.e. $$Z=V \cap U^c$$ for $$U,V \subset Y$$ open.

For any abelian sheaf $$\mathcal{F}$$ on $$Y$$ let $$\Gamma_Z(Y,\mathcal{F}):=\ker(\Gamma(V,\mathcal{F}) \rightarrow \Gamma(V-Z,\mathcal{F}))$$ be the sections of $$\mathcal{F}$$ with support in $$Z$$. Denoting by $$\mathfrak{Ab}(Y)$$ the category of abelian sheaves on $$Y$$, we have the functor $$\underline{\Gamma_Z}:\mathfrak{Ab}(Y) \rightarrow \mathfrak{Ab}(Y)$$, such that for $$\mathcal{F} \in \mathfrak{Ab}(Y)$$ and $$U\subset Y$$ open, we have $$\underline{\Gamma_Z}(\mathcal{F})(U)=\Gamma_{Z \cap U}(U,\mathcal{F}_{|U})$$.

Let $$\phi:(X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$$ be a flat morphism of locally noetherian locally ringed spaces, which induces the pullback $$\phi^*:\mathfrak{QCoh}(Y) \rightarrow \mathfrak{QCoh}(X)$$, where $$\mathfrak{QCoh}(X)$$ resp. $$\mathfrak{QCoh}(Y)$$ is the category of quasi-coherent sheaves on $$X$$ resp. $$Y$$. In this case we also have that $$\underline{\Gamma_Z}:\mathfrak{QCoh}(Y) \rightarrow \mathfrak{QCoh}(Y)$$.

Do we then have a commutative diagram

$$\require{AMScd}$$ $$\begin{CD} \mathfrak{QCoh}(Y) @>\underline{\Gamma_Z}>> \mathfrak{QCoh}(Y)\\ @V \phi^* V V @VV \phi^* V\\ \mathfrak{QCoh}(X) @>>\underline{\Gamma_{\phi^{-1}(Z)}}> \mathfrak{QCoh}(X) \end{CD}$$

?

• Is there a counter example if $\phi$ is not flat? Feb 10, 2020 at 5:25

Yes, for a flat morphism $$\phi:(X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$$ of quasi-compact and quasi-separated schemes one has an exact sequence $$0 \longrightarrow \underline{\Gamma_Z} \mathcal{F} \longrightarrow \mathcal{F} \longrightarrow j_*j^*\mathcal{F}$$ Where $$j \colon X \setminus Z \to X$$ denote the canonical embedding of the open complement. Applying $$\phi^*$$, one gets$$0 \longrightarrow \phi^*\underline{\Gamma_Z} \mathcal{F} \longrightarrow \phi^*\mathcal{F} \longrightarrow \phi^*j_*j^*\mathcal{F}$$ Let $$j' \colon Y \setminus Z' \to y$$ denote the corresponding canonical embedding with $$Z' := \phi^{-1}(Z)$$. We have the isomorphism $$\phi^*j_*j^*\mathcal{F} \cong j'_*j'^*\phi^* \mathcal{F}$$ Whence, as $$\underline{\Gamma_Z'}\phi^* \mathcal{F}$$ is the kernel of the map $$\phi^*\mathcal{F} \longrightarrow j'_*j'^*\phi^* \mathcal{F}$$ you obtain the desired identification $$\phi^*\underline{\Gamma_Z} \mathcal{F} \cong \underline{\Gamma_Z'}\phi^* \mathcal{F}.$$
• Just wondering, the isomorphism $\phi^*j_*j^*\mathcal{F}\cong j'_*j'^*\phi^*\mathcal{F}$ comes from flat base change, or? Does this isomorphisim also exists for the adic analytification $\phi: X^{ad} \rightarrow X$?
• It's flat base change plus pseudo-functoriality (transitivity) of $(-)_*$. You need some sort of base change for your $\phi$, to repeat the argument. Feb 13, 2020 at 13:30