Lemma 76.7.5. In Situation 76.7.1 suppose given an exact sequence

Then we have $F_{v, iso} = F_{u, zero}$ with obvious notation.

Lemma 76.7.5. In Situation 76.7.1 suppose given an exact sequence

\[ \mathcal{F} \xrightarrow {u} \mathcal{G} \xrightarrow {v} \mathcal{H} \to 0 \]

Then we have $F_{v, iso} = F_{u, zero}$ with obvious notation.

**Proof.**
Since pullback is right exact we see that $\mathcal{F}_ T \to \mathcal{G}_ T \to \mathcal{H}_ T \to 0$ is exact for every scheme $T$ over $B$. Hence $u_ T$ is surjective if and only if $v_ T$ is an isomorphism.
$\square$

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