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Proof. Let $E \subset X$ be locally constructible. We have to show that $f(E)$ is locally constructible too. We will show that $f(E) \cap V$ is constructible for any affine open $V \subset Y$. Thus we reduce to the case where $Y$ is affine. In this case $X$ is quasi-compact. Hence we can write $X = U_1 \cup \ldots \cup U_ n$ with each $U_ i$ affine open in $X$. If $E \subset X$ is locally constructible, then each $E \cap U_ i$ is constructible, see Properties, Lemma 28.2.1. Hence, since $f(E) = \bigcup f(E \cap U_ i)$ and since finite unions of constructible sets are constructible, this reduces us to the case where $X$ is affine. In this case the result is Algebra, Theorem 10.29.10. $\square$