Why the sheaf $j_{!}\mathcal{F}$ is zero outside $U$?

Question

Let $U\subset X$ be a oepn subset and $Z = X\setminus U$ is the complement,with inclusion $i:Z\to X$ and $j:U\to X$.

Let $\mathcal{F}_1$ be sheaf of module over $i^{-1}\mathscr{R}_X$ ($i^{-1}$ denote the inverse image) and $\mathcal{F}_2$ be sheaf of module over $j^{-1}\mathscr{R}_X$.

Then we define a category of triple:

With object $$\begin{aligned} &\left\{\left(\mathscr{F}_{1}, \mathscr{F}_{2}, \varphi\right) \mid \mathscr{F}_{1} \in \operatorname{Ob}\left(\mathscr{M} \operatorname{od}\left(i^{-1} \mathscr{R}_{X}\right)\right)\right. ,\left.\mathscr{F}_{2} \in \operatorname{Ob}\left(\mathscr{M} \operatorname{od}\left(j^{-1} \mathscr{R}_{X}\right)\right), \varphi: \mathscr{F}_{1} \rightarrow i^{-1} j_{*} \mathscr{F}_{2}\right\} \end{aligned}$$

And morphism $\left(\mathscr{F}_{1}, \mathscr{F}_{2}, \varphi\right) \rightarrow\left(\mathscr{F}_{1}^{\prime}, \mathscr{F}_{2}^{\prime}, \varphi^{\prime}\right)$ is a pair $\left(\psi_{1}, \psi_{2}\right)$ of morphisms $\psi_{1}: \mathscr{F}_{1} \rightarrow \mathscr{F}_{1}^{\prime}$ and $\psi_{2}: \mathscr{F}_{2} \rightarrow \mathscr{F}_{2}^{\prime}$ such that the diagram commute:

$$\begin{array}[c]{ccc} \mathcal{F}_1&\stackrel{\varphi}{\rightarrow}&i^{-1}j_*\mathcal{F_2}\\ \downarrow\scriptstyle{\psi_1}&&\downarrow\scriptstyle{i^{-1}j_*(\psi_2)}\\ \mathcal{F}_1'&\stackrel{\varphi'}{\rightarrow}&i^{-1}j_*\mathcal{F}_2' \end{array}$$

Define a functor $j_{!}:\mathcal{F_2}\mapsto (0,\mathcal{F_2},0)$ prove that the sheaf $j_{!}(\mathcal{F})$ has support on $Z$(i.e. zero outside $U$).

The key point is to check the stalk of $(0,\mathcal{F}_2,0)_x$ ,I don't know how to find the stalk for the triple $(0,\mathcal{F}_2,0)_x$?

There is another post define $j_{!}\mathcal{F}$ in an alternative way.that checking the stalk is a bit easier On the $j_!$ of a sheaf

Answer 1

Oh I found the solution,which prove the result as follows:Assume we already know the equivalence between sheaf of module and category of the triple.Denote this functor between them as $\Psi:\mathscr{Mod}_{Z, U}\left(\mathscr{R}_{X}\right)\to \mathscr{M}(\mathscr{R}_X)$(where $i:\mathscr{Mod}_{Z, U}\left(\mathscr{R}_{X}\right)$ be the category of the triple and $\mathscr{M}(\mathscr{R}_X)$ be the sheaf of the module).

Then we have following functor:

$$j_!:\mathcal{F}_2\to (0,\mathcal{F_2},0)\\j^{-1}\circ\Psi:(\mathcal{F}_1,\mathcal{F_2},\varphi)\to\mathcal{F_2}$$

Where $\mathcal{F}_2$ defined on space $U$.

Then easy to see the identity $j^{-1}\circ\Psi \circ j_! = id$,hence we can compute the stalk as follows:

$$\left(\Psi\circ j_{!} \mathscr{F}\right)_{x}=\left(\Psi\circ j_{!} \mathscr{F}\right)_{j(x)} \cong\left(j^{-1}\circ \Psi\circ j_{!} \mathscr{F}\right)_{x}=\mathscr{F}_{x}$$

For $x\in U$,for $x\in Z$ the result is similar.Hence we can prove that $j_!\mathcal{F}$ or more precisely the sheaf corresponding to it has stalk on $U$.

Moreover using the Cartesian diagram we can check that $$j^{-1}\circ\Psi :(\mathcal{F_1},\mathcal{F_2},\varphi) \mapsto \mathcal{F}_2$$ then use the same arguement as above ,we can show exactly the definition given by Roland