The Stacks project

Proof. The question is local on $Y$, hence we may assume that $Y$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ as a finite union of affine opens. Then $f' : X' = X_1 \amalg \ldots \amalg X_ n \to Y$ is a surjective flat morphism of affine schemes. Note that for $T \subset Y$ we have $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_ n$. Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open. Thus we may assume both $X$ and $Y$ are affine.

Let $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(J)$ for $J \subset B$ an ideal. Then $T = f(f^{-1}(T)) = f(V(J))$ is the image of $\mathop{\mathrm{Spec}}(B/J) \to \mathop{\mathrm{Spec}}(A)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 29.25.9). Hence by Topology, Lemma 5.19.6 we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under generalization. Hence $T$ is stable under specialization (Topology, Lemma 5.19.2). Thus $T$ is closed by Algebra, Lemma 10.41.5. $\square$