Lemma 67.28.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is of finite presentation and $Y$ is Noetherian, then $X$ is Noetherian.

**Proof.**
Assume $f$ is of finite presentation and $Y$ Noetherian. By Lemmas 67.28.5 and 67.23.5 we see that $X$ is locally Noetherian. As $f$ is quasi-compact and $Y$ is quasi-compact we see that $X$ is quasi-compact. As $f$ is of finite presentation it is quasi-separated (see Definition 67.28.1) and as $Y$ is Noetherian it is quasi-separated (see Properties of Spaces, Definition 66.24.1). Hence $X$ is quasi-separated by Lemma 67.4.9. Hence we have checked all three conditions of Properties of Spaces, Definition 66.24.1 and we win.
$\square$

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