**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

\[ \xymatrix{ Z \times _ Y X \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Z \ar[r]^ g & Y } \]

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.
$\square$

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