Lemma 108.4.5. Let B be an algebraic space. Let \pi : X \to Y be an open immersion of algebraic spaces which are separated and of finite presentation over B. Then the morphism \mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B} of Lemma 108.4.4 is an open immersion.
Proof. Omitted. Hint: If \mathcal{F} is an object of \mathcal{C}\! \mathit{oh}_{Y/B} over T and for t \in T we have \text{Supp}(\mathcal{F}_ t) \subset |X_ t|, then the same is true for t' \in T in a neighbourhood of t. \square
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