The Stacks project

Proof. For every $a \in A$ and $b \in B$ there exist opens $U_{(a, b)}$ of $X$ and $V_{(a, b)}$ of $Y$ such that $(a, b) \in U_{(a, b)} \times V_{(a, b)} \subset W$. Fix $b$ and we see there exist a finite number $a_1, \ldots , a_ n$ such that $A \subset U_{(a_1, b)} \cup \ldots \cup U_{(a_ n, b)}$. Hence

Thus for every $b \in B$ there exists opens $U_ b \subset X$ and $V_ b \subset Y$ such that $A \times \{ b\} \subset U_ b \times V_ b \subset W$. As above there exist a finite number $b_1, \ldots , b_ m$ such that $B \subset V_{b_1} \cup \ldots \cup V_{b_ m}$. Then we win because $A \times B \subset (U_{b_1} \cap \ldots \cap U_{b_ m}) \times (V_{b_1} \cup \ldots \cup V_{b_ m})$. $\square$

Proof. (See also remark below.) If $X$ is not quasi-compact, there exists an open covering $X = \bigcup _{i \in I} U_ i$ such that no finite number of $U_ i$ cover $X$. Let $Z$ be the subset of the power set $\mathcal{P}(I)$ of $I$ consisting of $I$ and all nonempty finite subsets of $I$. Define a topology on $Z$ with as a basis for the topology the following sets:

1. All subsets of $Z\setminus \{ I\} $.

2. For every finite subset $K$ of $I$ the set $U_ K := \{ J\subset I \mid J \in Z, \ K\subset J \} )$.

It is left to the reader to verify this is the basis for a topology. Consider the subset of $Z \times X$ defined by the formula

If $(J, x) \not\in M$, then $x \in U_ i$ for some $i \in J$. Hence $U_{\{ i\} } \times U_ i \subset Z \times X$ is an open subset containing $(J, x)$ and not intersecting $M$. Hence $M$ is closed. The projection of $M$ to $Z$ is $Z-\{ I\} $ which is not closed. Hence $Z \times X \to Z$ is not closed.

Assume $X$ is quasi-compact. Let $Z$ be a topological space. Let $M \subset Z \times X$ be closed. Let $z \in Z$ be a point which is not in $\text{pr}_1(M)$. By the Tube Lemma 5.17.1 there exists an open $U \subset Z$ such that $U \times X$ is contained in the complement of $M$. Hence $\text{pr}_1(M)$ is closed. $\square$

Proof. (See also the remark below.) If the map $f$ satisfies (1), it automatically satisfies (4) because any single point is quasi-compact.

Assume map $f$ satisfies (4). We will prove it is universally closed, i.e., (3) holds. Let $g : Z \to Y$ be a continuous map of topological spaces and consider the diagram

During the proof we will use that $Z \times _ Y X \to Z \times X$ is a homeomorphism onto its image, i.e., that we may identify $Z \times _ Y X$ with the corresponding subset of $Z \times X$ with the induced topology. The image of $f' : Z \times _ Y X \to Z$ is $\mathop{\mathrm{Im}}(f') = \{ z : g(z) \in f(X)\} $. Because $f(X)$ is closed, we see that $\mathop{\mathrm{Im}}(f')$ is a closed subspace of $Z$. Consider a closed subset $P \subset Z \times _ Y X$. Let $z \in Z$, $z \not\in f'(P)$. If $z \not\in \mathop{\mathrm{Im}}(f')$, then $Z \setminus \mathop{\mathrm{Im}}(f')$ is an open neighbourhood which avoids $f'(P)$. If $z$ is in $\mathop{\mathrm{Im}}(f')$ then $(f')^{-1}\{ z\} = \{ z\} \times f^{-1}\{ g(z)\} $ and $f^{-1}\{ g(z)\} $ is quasi-compact by assumption. Because $P$ is a closed subset of $Z \times _ Y X$, we have a closed $P'$ of $Z \times X$ such that $P = P' \cap Z \times _ Y X$. Since $(f')^{-1}\{ z\} $ is a subset of $P^ c = P'^ c \cup (Z \times _ Y X)^ c$, and since $(f')^{-1}\{ z\} $ is disjoint from $(Z \times _ Y X)^ c$ we see that $(f')^{-1}\{ z\} $ is contained in $P'^ c$. We may apply the Tube Lemma 5.17.1 to $(f')^{-1}\{ z\} = \{ z\} \times f^{-1}\{ g(z)\} \subset (P')^ c \subset Z \times X$. This gives $V \times U$ containing $(f')^{-1}\{ z\} $ where $U$ and $V$ are open sets in $X$ and $Z$ respectively and $V \times U$ has empty intersection with $P'$. Then the set $V \cap g^{-1}(Y-f(U^ c))$ is open in $Z$ since $f$ is closed, contains $z$, and has empty intersection with the image of $P$. Thus $f'(P)$ is closed. In other words, the map $f$ is universally closed.

The implication (3) $\Rightarrow $ (2) is trivial. Namely, given any topological space $Z$ consider the projection morphism $g : Z \times Y \to Y$. Then it is easy to see that $f'$ is the map $Z \times X \to Z \times Y$, in other words that $(Z \times Y) \times _ Y X = Z \times X$. (This identification is a purely categorical property having nothing to do with topological spaces per se.)

Assume $f$ satisfies (2). We will prove it satisfies (1). Note that $f$ is closed as $f$ can be identified with the map $\{ pt\} \times X \to \{ pt\} \times Y$ which is assumed closed. Choose any quasi-compact subset $K \subset Y$. Let $Z$ be any topological space. Because $Z \times X \to Z \times Y$ is closed we see the map $Z \times f^{-1}(K) \to Z \times K$ is closed (if $T$ is closed in $Z \times f^{-1}(K)$, write $T = Z \times f^{-1}(K) \cap T'$ for some closed $T' \subset Z \times X$). Because $K$ is quasi-compact, $K \times Z\to Z$ is closed by Lemma 5.17.3. Hence the composition $Z \times f^{-1}(K)\to Z \times K \to Z$ is closed and therefore $f^{-1}(K)$ must be quasi-compact by Lemma 5.17.3 again. $\square$

Proof. Since every point of $Y$ is closed, we see from Lemma 5.12.3 that the closed subset $f^{-1}(y)$ of $X$ is quasi-compact for all $y \in Y$. Thus, by Theorem 5.17.5 it suffices to show that $f$ is closed. If $E \subset X$ is closed, then it is quasi-compact (Lemma 5.12.3), hence $f(E) \subset Y$ is quasi-compact (Lemma 5.12.7), hence $f(E)$ is closed in $Y$ (Lemma 5.12.4). $\square$

Proof. It suffices to prove $f$ is closed, because this implies that $f^{-1}$ is continuous. If $T \subset X$ is closed, then $T$ is quasi-compact by Lemma 5.12.3, hence $f(T)$ is quasi-compact by Lemma 5.12.7, hence $f(T)$ is closed by Lemma 5.12.4. $\square$