3
$\begingroup$

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}

?

$\endgroup$
1
  • $\begingroup$ Is there a counter example if $\phi$ is not flat? $\endgroup$ – Karl Schwede Feb 10 '20 at 5:25
3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 $\phi: X^{ad} \rightarrow X$? $\endgroup$ – KKD Feb 13 '20 at 10:54
  • $\begingroup$ It's flat base change plus pseudo-functoriality (transitivity) of $(-)_*$. You need some sort of base change for your $\phi$, to repeat the argument. $\endgroup$ – Leo Alonso Feb 13 '20 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.