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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