Lemma 81.10.2. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset X$ closed subspace such that $f^{-1}Z \to Z$ is integral and universally injective. Let $\overline{y}$ be a geometric point of $Y$ and $\overline{x} = f(\overline{y})$. Let $\mathcal{G}$ be an abelian sheaf on $Y$. Then the map of two term complexes
\[ \left(f_*\mathcal{G}_{\overline{x}} \to (f \circ j')_*(\mathcal{G}|_ V)_{\overline{x}}\right) \longrightarrow \left(\mathcal{G}_{\overline{y}} \to j'_*(\mathcal{G}|_ V)_{\overline{y}}\right) \]
induces an isomorphism on kernels and an injection on cokernels. Here $V = Y \setminus f^{-1}Z$ and $j' : V \to Y$ is the inclusion.
Proof.
Choose a distinguished triangle
\[ \mathcal{G} \to Rj'_*\mathcal{G}|_ V \to Q \to \mathcal{G}[1] \]
n $D(Y_{\acute{e}tale})$. The cohomology sheaves of $Q$ are supported on $|f^{-1}Z|$. We apply $Rf_*$ and we obtain
\[ Rf_*\mathcal{G} \to Rf_*Rj'_*\mathcal{G}|_ V \to Rf_*Q \to Rf_*\mathcal{G}[1] \]
Taking stalks at $\overline{x}$ we obtain an exact sequence
\[ 0 \to (R^{-1}f_*Q)_{\overline{x}} \to f_*\mathcal{G}_{\overline{x}} \to (f \circ j')_*(\mathcal{G}|_ V)_{\overline{x}} \to (R^0f_*Q)_{\overline{x}} \]
We can compare this with the exact sequence
\[ 0 \to H^{-1}(Q)_{\overline{y}} \to \mathcal{G}_{\overline{y}} \to j'_*(\mathcal{G}|_ V)_{\overline{y}} \to H^0(Q)_{\overline{y}} \]
Thus we see that the lemma follows because $Q_{\overline{y}} = Rf_*Q_{\overline{x}}$ by Lemma 81.10.1.
$\square$
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