Improve a proof

Thanks to Keenan Kidwell
http://stacks.math.columbia.edu/tag/08KS#comment-1855

modules.tex

@@ -2033,23 +2033,27 @@ \section{Closed immersions of ringed spaces}
 
 \medskip\noindent
 Let $\mathcal{G}$ be a $\mathcal{O}_X$-module with
-$\mathcal{I}\mathcal{G} = 0$. If $x \in X$, $x \not \in i(Z)$, then
-$\mathcal{G}_x = 0$ because $\mathcal{I}_x = \mathcal{O}_{X, x}$ in this
-case. Thus we see that $\mathcal{G}$ us supported on $Z$.
-By Lemma \ref{lemma-i-star-exact} we can write $\mathcal{G} = i_*\mathcal{F}$
-for a unique abelian sheaf $\mathcal{F}$ on $Z$.
-Let $W \subset Z$ be open, $f \in \mathcal{O}_Z(W)$ and $s \in \mathcal{F}(W)$.
-We define $fs \in \mathcal{F}(W)$. Since $i^\sharp$ is surjective
-we can find opens $U_j \subset X$ such that $W = \bigcup i^{-1}(U_j)$
-and $f|{i^{-1}(U_j)}$ is the image of $f_j \in \mathcal{O}_X(U_j)$.
-Note that $s|_{i^{-1}(U_j)}$ is an element of
-$\mathcal{F}(i^{-1}(U_j)) = \mathcal{G}(U_i)$.
-Thus we can form $s_j = f_js \in \mathcal{F}(i^{-1}(U_j)) = \mathcal{G}(U_i)$.
-By our assumption that $\mathcal{I}\mathcal{G} = 0$
-the sections $s_j$ are independent of the choice of $f_j$ lifting
-$f|{i^{-1}(U_j)}$ and glue to a section $fs$ of $\mathcal{F}$ over $W$.
-In this way $\mathcal{F}$ becomes an $\mathcal{O}_Z$-module
-such that $\mathcal{G} \cong i_*\mathcal{F}$.
+$\mathcal{I}\mathcal{G} = 0$. We will prove the canonical map
+$$
+\mathcal{G} \longrightarrow i_*i^*\mathcal{G}
+$$
+is an isomorphism. This proves that $\mathcal{G} = i_*\mathcal{F}$
+with $\mathcal{F} = i^*\mathcal{G}$ which finishes the proof.
+We check the displayed map induces an isomorphism on stalks.
+If $x \in X$, $x \not \in i(Z)$, then $\mathcal{G}_x = 0$
+because $\mathcal{I}_x = \mathcal{O}_{X, x}$ in this
+case. As above $(i_*i^*\mathcal{G})_x = 0$ by
+Sheaves, Lemma \ref{sheaves-lemma-stalks-closed-pushforward}.
+On the other hand, if $x \in Z$, then we obtain the map
+$$
+\mathcal{G}_x
+\longrightarrow
+\mathcal{G}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Z, x}
+$$
+by Sheaves, Lemmas \ref{sheaves-lemma-stalk-pullback-modules} and
+\ref{sheaves-lemma-stalks-closed-pushforward}. This map is an isomorphism
+because $\mathcal{O}_{Z, x} = \mathcal{O}_{X, x}/\mathcal{I}_x$
+and because $\mathcal{G}_x$ is annihilated by $\mathcal{I}_x$ by assumption.
 \end{proof}