Does local cohomology commute with pullback?

Question

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}

?

Answer 1

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 $$ \phi^*\underline{\Gamma_Z} \mathcal{F} \cong \underline{\Gamma_Z'}\phi^* \mathcal{F}. $$