Lemma 5.17.1 (Tube lemma). Let $X$ and $Y$ be topological spaces. Let $A \subset X$ and $B \subset Y$ be quasi-compact subsets. Let $A \times B \subset W \subset X \times Y$ with $W$ open in $X \times Y$. Then there exists opens $A \subset U \subset X$ and $B \subset V \subset Y$ such that $U \times V \subset W$.

## 5.17 Characterizing proper maps

We include a section discussing the notion of a proper map in usual topology. It turns out that in topology, the notion of being proper is the same as the notion of being universally closed, in the sense that any base change is a closed morphism (not just taking products with spaces). The reason for doing this is that in algebraic geometry we use this notion of universal closedness as the basis for our definition of properness.

**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 $A \times B \subset (U_{b_1} \cap \ldots \cap U_{b_ m}) \times (V_{b_1} \cup \ldots \cup V_{b_ m})$. $\square$

The notation in the following definition may be slightly different from what you are used to.

Definition 5.17.2. Let $f : X\to Y$ be a continuous map between topological spaces.

We say that the map $f$ is

*closed*iff the image of every closed subset is closed.We say that the map $f$ is

*proper*^{1}iff the map $Z \times X\to Z \times Y$ is closed for any topological space $Z$.We say that the map $f$ is

*quasi-proper*iff the inverse image $f^{-1}(V)$ of every quasi-compact subset $V \subset Y$ is quasi-compact.We say that $f$ is

*universally closed*iff the map $f': Z \times _ Y X \to Z$ is closed for any map $g: Z \to Y$.

The following lemma is useful later.

Lemma 5.17.3. A topological space $X$ is quasi-compact if and only if the projection map $Z \times X \to Z$ is closed for any topological space $Z$.

**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:

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

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. $\square$

Remark 5.17.4. Lemma 5.17.3 is a combination of [I, p. 75, Lemme 1, Bourbaki] and [I, p. 76, Corollaire 1, Bourbaki].

Theorem 5.17.5. Let $f: X\to Y$ be a continuous map between topological spaces. The following conditions are equivalent:

The map $f$ is quasi-proper and closed.

The map $f$ is proper.

The map $f$ is universally closed.

The map $f$ is closed and $f^{-1}(y)$ is quasi-compact for any $y\in Y$.

**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. $\square$

Remark 5.17.6. Here are some references to the literature. In [I, p. 75, Theorem 1, Bourbaki] you can find: (2) $\Leftrightarrow $ (4). In [I, p. 77, Proposition 6, Bourbaki] you can find: (2) $\Rightarrow $ (1). Of course, trivially we have (1) $\Rightarrow $ (4). Thus (1), (2) and (4) are equivalent. Fan Zhou claimed and proved that (3) and (4) are equivalent; let me know if you find a reference in the literature.

Lemma 5.17.7. Let $f : X \to Y$ be a continuous map of topological spaces. If $X$ is quasi-compact and $Y$ is Hausdorff, then $f$ is proper.

**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).
$\square$

Lemma 5.17.8. Let $f : X \to Y$ be a continuous map of topological spaces. If $f$ is bijective, $X$ is quasi-compact, and $Y$ is Hausdorff, then $f$ is a homeomorphism.

**Proof.**
This follows immediately from Lemma 5.17.7 which tells us that $f$ is closed, i.e., $f^{-1}$ is continuous.
$\square$

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