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Let $X$ be a scheme and $j:Z\subseteq X$ be a closed subscheme. Let $\mathcal{F}$ be a sheaf of modules over $X$. Show that \begin{equation*} j_{*}j^{-1}\mathcal{F}_{x}=\left\{\begin{matrix}\mathcal{F}_{x} & \text{ if $x\in Z$}\\0 & \text{otherwise}\end{matrix}\right. \end{equation*} I think I can show that $j_{*}j^{-1}\mathcal{F}_{x}=0$ if $x\notin Z$ and $j_{*}j^{-1}\mathcal{F}_{x}=\mathcal{F}_{x}$ if $x$ is contained in $Z^{o}$ (the largest open subset of $Z$), but what do I do if $x\in Z\setminus Z^{o}$?

The Thin Whistler
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  • Are you aware of what $j^{-1}$ and $j_!$ do to stalks in general? If not, that might be a good thing to find out! – KReiser Aug 25 '20 at 18:17
  • @KReiser can you give me a hint? Or a reference? I mean $j_{!}$ does not change stalks, if I am not mistaken. But what about $j^{-1}$? – The Thin Whistler Aug 25 '20 at 18:18
  • Well, what would you want it to do? What $j^{-1}$ does to stalks has been covered on this website before, it's covered on many other websites (wikipedia, etc - try searching), and you can prove it yourself. – KReiser Aug 25 '20 at 18:31
  • I think I got it now: $j_{*}j^{-1}F(V)=\mathcal{F}(i(i^{-1}(V)))$, so the answer is simple. Am I right? – The Thin Whistler Aug 25 '20 at 19:02
  • No, that's very wrong. $(j^{-1}F)(U)$ is the limit of $F(V)$ over all $V$ containing $j(U)$, which makes your claim in the previous comment fail horrendously unless $j(U)$ is open (it almost never is). I've already told you that what $j^{-1}$ does to stalks is well-documented on this website and on wikipedia - search "inverse image stalks". Good day! – KReiser Aug 25 '20 at 19:48
  • you are right, got it. – The Thin Whistler Aug 25 '20 at 20:16

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