Lemma 66.16.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. There exists a closed subspace $Z \subset Y$ such that $f$ factors through $Z$ and such that for any other closed subspace $Z' \subset Y$ such that $f$ factors through $Z'$ we have $Z \subset Z'$.

** The scheme-theoretic image of a morphism of algebraic spaces exists. **

**Proof.**
Let $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$. If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$, see Lemma 66.13.1. In general the lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$. This follows from Lemma 66.14.2.
$\square$

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## Comments (1)

Comment #1285 by Johan Commelin on