# The Stacks Project

## Tag 02JY

Lemma 28.24.11. Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism. A subset $T \subset Y$ is open (resp. closed) if and only $f^{-1}(T)$ is open (resp. closed). In other words, $f$ is a submersive morphism.

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 $f : \mathop{\rm Spec}(B) \to \mathop{\rm Spec}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(I)$ for $I \subset A$ an ideal. Then $T = f(f^{-1}(T)) = f(V(I))$ is the image of $\mathop{\rm Spec}(A/I) \to \mathop{\rm Spec}(B)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 28.24.8). Hence by Topology, Lemma 5.19.5 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 10.40.5. $\square$

The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4382–4388 (see updates for more information).

\begin{lemma}
\label{lemma-fpqc-quotient-topology}
Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism.
A subset $T \subset Y$ is open (resp.\ closed) if and only
$f^{-1}(T)$ is open (resp.\ closed). In other words, $f$ is
a submersive morphism.
\end{lemma}

\begin{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.

\medskip\noindent
Let $f : \Spec(B) \to \Spec(A)$ be a surjective
morphism of affine schemes corresponding to a flat ring map $A \to B$.
Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(I)$ for $I \subset A$
an ideal. Then $T = f(f^{-1}(T)) = f(V(I))$ is the image of
$\Spec(A/I) \to \Spec(B)$ (here we use that $f$
is surjective). On the other hand, generalizations lift along $f$
(Lemma \ref{lemma-generalizations-lift-flat}).
Hence by Topology, Lemma \ref{topology-lemma-lift-specializations-images}
we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under
generalization. Hence $T$ is stable under specialization
(Topology, Lemma \ref{topology-lemma-open-closed-specialization}).
Thus $T$ is closed by
Algebra, Lemma \ref{algebra-lemma-image-stable-specialization-closed}.
\end{proof}

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