Lemma 108.10.3. With $B, X, Y$ as in the introduction of this section, in addition assume $X \to B$ is proper. Then the subfunctor $\mathit{Isom}_ B(Y, X) \subset \mathit{Mor}_ B(Y, X)$ of isomorphisms is an open subspace.
Proof. Follows immediately from More on Morphisms of Spaces, Lemma 76.49.6. $\square$
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