We have just seen that a) implies b). Since the equivalence relation $f(x) = f(x')$ is the equality relation, the quotient space of $X$ with respect to this relation can be identified with $X$; hence b) implies c) by reason of § 5, no. 2, Proposition 3. Finally, if c) is satisfied then $f \times \iota_Z$ is a homeomorphism of $X \times Z$ onto a closed subspace of $Y \times Z$ and is therefore a closed mapping; hence c) implies a).

Let $Z$ be a topological space. If $T$ is any subset of $Y$, we have

if $f$ is proper, then $f \times \iota_Z$ is closed, hence so is $(f \times \iota_Z)_{T \times Z}$ [§ 5, no. 1, Proposition 2 a)], whence a) is proved. If now $(T(i))_{i \in I}$ satisfies one of the two conditions stated in b), then the covering $(T(i) \times Z)_{i \in I}$ of $Y \times Z$ has the same property; if the $f_{T(i)}$ are proper then the mappings are closed, whence $f \times \iota_Z$ is closed [§ 5, no. 1, Proposition 2 b)]. This completes the proof.

(We shall see in no. 2, Theorem 1, Corollary 3, that this proposition extends to infinite products.)

By induction it is enough to consider the case where $I = \{1, 2\}$. a) Suppose that $f_1, f_2$ are proper, and let $Z$ be a topological space; $f_1 \times f_2 \times \iota_Z$ is the composition of $\iota_{Y_1} \times f_2 \times \iota_Z$ and

these two mappings are closed by hypothesis, hence so is $f_1 \times f_2 \times \iota_Z$ [§ 5, no. 1, Proposition 1 a)], whence $f_1 \times f_2$ is proper.

b) Now suppose $f$ is proper. Let $F$ be a closed subset of $X_2 \times Z$ and let $G$ be the image of $F$ in $Y_2 \times Z$ under the mapping $f_2 \times \iota_Z$. Then the image of $X_1 \times F$ in $Y_1 \times Y_2 \times Z$ under $f_1 \times f_2 \times \iota_Z$ is $f_1(X_1) \times G$. By hypothesis, this is closed in $Y_1 \times Y_2 \times Z$; if $X_1 \neq \emptyset$, then $f_1(X_1)$ is not empty, which implies that $G$ is closed in $Y_2 \times Z$ (§ 4, no. 3, Corollary to Proposition 7); hence $f_2$ is proper. Similarly $f_1$ is proper if $X_2 \neq \emptyset$.

Let $Z$ be a topological space. We have

if $f$ and $g$ are proper, then $f \times \iota_Z$ and $g \times \iota_Z$ are closed; hence [§ 5, no. 1, Proposition 1 a)] $(g \circ f) \times \iota_Z$ is closed; this proves a). The proof of b) [resp. c)] runs along the same lines, using part b) [resp. c)] of Proposition 1 of § 5, no. 1, and remarking that if $f$ is surjective (resp. if $g$ is injective) then $f \times \iota_Z$ is surjective (resp. $g \times \iota_Z$ is injective). Finally, to prove d), consider the commutative diagram

where $\varphi(x) = (x, f(x))$ and $\psi(x') = (g(x'), x')$. The mapping $\varphi$ (resp. $\psi$) is a homeomorphism of $X$ (resp. $X'$) onto the graph of $f$ (resp. the reflection of the graph of $g$) (§ 4, no. 1, Proposition 1, Corollary 2). Further, since $X'$ is Hausdorff, the graph $\varphi(X)$ of $f$ is closed in $X \times X'$ (§ 8, no. 1, Proposition 2, Corollary 2). Hence (Proposition 2) $\varphi$ is proper; on the other hand Proposition 4 shows that $(g \circ f) \times i_{X'}$ is proper. By a) above and the commutativity of the diagram (1), $\psi \circ f$ is proper; but $\psi$ is injective and therefore $f$ is proper by c) above.

For this restriction is the composition $f \circ j$, where $j : F \to X$ is the canonical injection, which is proper by Proposition 2.

By reason of Proposition 5 c) we need only consider the case where $f(X) = Y$. Then the diagonal of $Y \times Y$ is the image under $f \times f$ of the diagonal of $X$, which is closed (§ 8, no. 1, Proposition 1); $f \times f$ is proper (Proposition 4); hence the diagonal of $Y \times Y$ is closed (Proposition 1) and therefore $Y$ is Hausdorff (§ 8, no. 1, Proposition 1).

This mapping is the composition of the product mapping $(x_i) \mapsto (f_i(x_i))$ of $X^I$ into $\prod_i Y_i$ and the diagonal mapping of $X$ into $X^I$; since the latter is proper (by Proposition 2 and § 8, no. 1, Proposition 1) the conclusion follows from Proposition 4 and Proposition 5 a).

The conditions are sufficient by virtue of Proposition 5 a) and Proposition 2. Conversely, if $f$ is proper, then $f$ is closed; hence $f(X)$ is closed in $Y$ and $h$ is a homeomorphism ($§ 5$, no. 2, Proposition 3); also $h \circ p$ is proper by Proposition 5 c); hence $p = h^{-1} \circ (h \circ p)$ is proper by Proposition 5 a).

(We shall see a little later on (Theorem 1, Corollary 1) that this property characterizes quasi-compact spaces.)

We may restrict ourselves to the case where $X$ is not empty. Let $\mathfrak{F}$ be a filter on $X$, and $X' = X \cup \{ \omega \}$ the topological space associated with $\mathfrak{F}$ ($§ 6$, no. 5, Example). Let $\Delta$ be the subset of $X \times X'$ consisting of all $(x, x)$ where $x \in X$, and let $F = \overline{\Delta}$ be the closure of $\Delta$ in $X \times X'$. In view of the hypothesis on $X$, the image of $F$ under the projection $X \times X' \to X'$ is closed in $X'$; this image contains $X$ and therefore contains $\omega$, which lies in the closure of $X$; in other words, there is a point $x \in X$ such that $(x, \omega) \in F$. By the definition of the topology of $X \times X'$, this means that, for each neighbourhood $V$ of $x$ in $X$ and each $M \in \mathfrak{F}$, we have $(V \times M) \cap \Delta \neq \emptyset$, i.e. $V \cap M \neq \emptyset$, so that $x$ is a cluster point of the filter $\mathfrak{F}$, and therefore $X$ is quasi-compact.

Q.E.D.

a) $\Longrightarrow$ b): If $f$ is proper then $f$ is closed (no. 1, Proposition 1) and for each $y \in Y$ the mapping $f_{|y|} : \overline{f}^{-1}(y) \to \{ y \}$ is proper [no. 1, Proposition 3a)]. By Lemma 1, this implies that $\overline{f}^{-1}(y)$ is quasi-compact.

b) $\Longrightarrow$ c): Suppose $\mathfrak{F}$ and $y$ satisfy the hypotheses of c). Let $\mathfrak{B}$ be the filter base on $X$ formed by the closures of the sets of $\mathfrak{F}$. Since $f$ is closed, we have $f(\overline{M}) = f(M)$ for each $M \in \mathfrak{F}$ (§ 5, no. 4, Proposition 9). This shows that the sets $\overline{M} \cap \overline{f}^{-1}(y)$ are non-empty for all $M \in \mathfrak{F}$, and hence form a filter base on $\overline{f}^{-1}(y)$ whose elements are closed subsets of $\overline{f}^{-1}(y)$. Since $\overline{f}^{-1}(y)$ is quasi-compact, there is a point $x \in \overline{f}^{-1}(y)$ which belongs to all the sets $M$ as $M$ runs through $\mathfrak{F}$. Hence $f(x) = y$ and $x$ is a cluster point of $\mathfrak{F}$.

c) $\Longrightarrow$ d): Trivial.

d) $\Longrightarrow$ a): We show first that if d) is satisfied, then $f$ is a closed mapping. Let $A$ be a non-empty closed subset of $X$ and let $\mathfrak{F}$ be the filter of subsets of $X$ which contain $A$. Then $A$ is the set of cluster points of $\mathfrak{F}$. Let $B$ be the set of cluster points of the filter base $f(\mathfrak{F})$ on $Y$; $B$ is closed and clearly contains $f(A)$; we shall show that $B = f(A)$. Let $y \in B$ and let $\mathfrak{B}$ be the neighbourhood filter of $y$ in $Y$; then by hypothesis every set of $\mathfrak{W} = \overline{f}^{-1}(\mathfrak{B})$ meets every set of $\mathfrak{F}$; hence $\mathfrak{W}$ is a filter base on $X$ and there is an ultrafilter $U$ on $X$ which is finer than both $\mathfrak{F}$ and the filter whose base is $\mathfrak{W}$ (§ 6, no. 2, Proposition 1, Corollary 1 and no. 4, Theorem 1). The ultrafilter whose base is $f(U)$ is finer than $\mathfrak{B}$ and therefore converges to $y$. By virtue of d) there is a point $x \in X$ such that $f(x) = y$ and $U$ converges to $x$; since $U$ is finer than $\mathfrak{F}$, $x$ is a cluster point of $\mathfrak{F}$; hence $x \in A$. This shows that $B = f(A)$ and therefore that $f$ is closed.

To complete the proof we have to show that $f \times \iota_Z$ is closed for every topological space $Z$. From what has been proved it is enough to show that if $f$ satisfies condition d), then so does $f \times \iota_Z$. This is a consequence of the following general lemma:

Let $U$ be an ultrafilter on $X = \prod_i X_i$, and let $y = (y_i)$ be a point of $Y = \prod_i Y_i$ such that $f(U)$ converges to $y$. This means that each of the ultrafilter bases $\operatorname{pr}_i(f(U)) = f_i(\operatorname{pr}_i(U))$ converges to $y_i$ (§ 7, no. 6, Proposition 10, Corollary 1). By virtue of condition d), for each $i \in I$ there exists $x_i \in X_i$ such that $f_i(x_i) = y_i$ and $\operatorname{pr}_i(U)$ converges to $x_i$; but then $U$ converges to $x = (x_i)$ (loc. cit.) and we have $f(x) = y$. This completes the proof of Lemma 2 and hence of Theorem 1.

Apply a) $\iff$ b) to $X \to P$.

The composition $X \xrightarrow{f} Y \to P$ is proper by Corollary 1; hence $f$ is proper by no. 1, Proposition 5 d). Alternatively we may apply the criterion b) of Theorem 1, using § 9, no. 4, Theorem 2, Corollary 2.

In view of Theorem 1, this is just Lemma 2 above.

If we apply this corollary to the family of mappings $X_i \to P$ and use Corollary 1, we have Tychonoff’s theorem (§ 9, no 5, Theorem 3).

The proof is the same as in the case of a finite family (no. 1, Proposition 5, Corollary 3), using Corollary 3 above and the fact that the diagonal of $X^1$ is closed (§ 8, no. 1, Proposition 1).

For we may identify $Y$ with $P \times Y$ and $\mathrm{pr}_2$ with the product of $X \to P$ and $\iota_Y$, both of which are proper mappings.

By Proposition 3 of no. 1, the mapping $f_K : \overline{f}^{-1}(K) \to K$ is proper. Since $K \to P$ is a proper mapping (Theorem 1, Corollary 1) it follows from no. 1, Proposition 5 a) that the composition $\overline{f}^{-1}(K) \xrightarrow{f_K} K \to P$ is proper, whence $\overline{f}^{-1}(K)$ is quasi-compact by Theorem 1, Corollary 1.

If f is proper and K is a compact subset of Y, then $\overline{f}^{-1}(K)$ is compact by Proposition 6 of no. 2. Conversely, if this condition is satisfied, let $(\mathbf{U}_\alpha)$ be a covering of Y by relatively compact open sets. Then the sets $\overline{f}^{-1}(\overline{\mathbf{U}}_\alpha)$ are compact in X and their interiors cover X; since X is Hausdorff this shows that X is locally compact. Furthermore, each of the mappings $f_{\overline{\mathbf{U}}_\alpha} : \overline{f}^{-1}(\alpha \overline{\mathbf{U}}) \to \overline{\mathbf{U}}_\alpha$ is proper (no. 2, Theorem 1, Corollary 2) and therefore f is proper by Proposition 3 b) of no. 1.

By Proposition 7, f is proper if and only if, for each compact subset K' of X', $\overline{f}^{-1}(X' - K') = X - \overline{f}^{-1}(K')$ is the complement of a compact subset of X; by the definition of the neighbourhoods of $\omega$ (resp. $\omega'$) in Y (resp. Y') this is so if and only if $\overline{f}$ is continuous at $\omega$.

a) $C$ is closed in $X \times X$.

b) $R$ is closed.

c) $f$ is proper.

d) $X/R$ is Hausdorff.

Furthermore, if these conditions are satisfied then $X/R$ is compact.

$R$ is closed if and only if $f$ is closed; hence b) implies c) by reason of Theorem 1b) of no. 2. That c) implies d) is a particular case of no. 1, Proposition 5, Corollary 2. d) implies a) for any topological space $X$ ($\S$ 8, no. 3, Proposition 8). It remains to show that a) implies b). If $F$ is a closed subset of $X$ its saturation (with respect to $R$) is $\mathrm{pr}_2(C \cap (F \times X))$; by hypothesis, $C \cap (F \times X)$ is a closed subset of the compact space $X \times X$, and is therefore compact ($\S$ 9, no. 3, Proposition 3); the result now follows from the continuity of $\mathrm{pr}_2$ ($\S$ 9, no. 4, Theorem 2, Corollary 2).

Finally it is clear that if $X/R$ is Hausdorff then it is compact ($\S$ 9, no. 4, Theorem 2).

a) $f$ is proper.

b) The saturation of each compact subset of $X$ with respect to $R$ is compact.

c) $R'$ is closed.

d) The restriction of $\mathrm{pr}_2$ to $C$ is proper.

e) $R$ is closed and the equivalence classes with respect to $R$ are compact.

Furthermore, if these conditions are satisfied, $X/R$ is locally compact.

a) $\Longrightarrow$ b): Since $X/R = f(X)$ and $f$ is proper, $X/R$ is Hausdorff (no. 1, Proposition 5, Corollary 2); hence the image under $f$ of every compact subset $K$ of $X$ is compact ($\S$ 9, no. 4, Theorem 2, Corollary 1).

The saturation of $K$ with respect to $R$ is $\overline{f}^{-1}(f(K))$ and is therefore compact by no. 2, Proposition 6.

b) $\Longrightarrow$ c) : If $F'$ is closed in $X'$ and does not contain $\omega$, then $F'$ is a compact subset of $X$; hence its saturation with respect to $R'$, which is the same as its saturation with respect to $R$, is compact and a fortiori closed in $X'$. If $\omega \in F'$ and if $F = F' \cap X = F' - \{ \omega \}$, then the saturation of $F'$ with respect to $R'$ is the union of $\{ \omega \}$ and the saturation $H$ of $F$ with respect to $R$; hence it is enough to show that $H$ is closed in $X$ (i.e. that $R$ is a closed relation). For this, it is enough to show that if $K$ is any compact subset of $X$ then $H \cap K$ is compact ($§ 9$, no. 7, Proposition 11). Now the saturation $L$ of $K$ with respect to $R$ is compact by hypothesis, and $H \cap L$ is the saturation of $F \cap L$, which is also compact; a fortiori $H \cap K = (H \cap L) \cap K$ is compact.

c) $\Longrightarrow$ d) : Since $X'$ is regular ($§ 9$, no. 2, Corollary to Proposition 1), $C'$ is closed in $X' \times X'$ ($§ 8$, no. 6, Proposition 14) and therefore compact. It follows that $C'$ is the one-point compactification of $C$ ($§ 9$, no. 8, Theorem 4). Since the restriction to $C'$ of $pr_2 : X' \times X' \to X'$ is continuous at $\omega$, the result follows from no. 3, Corollary to Proposition 7.

d) $\Longrightarrow$ e) : If $F$ is any closed subset of $X$, then $C \cap (F \times X)$ is closed in $C$, whence the saturation of $F$ with respect to $R$, which is equal to $pr_2(C \cap (F \times X))$, is closed in $X$ (no. 1, Proposition 1). Also the equivalence class of $x \in X$ mod $R$ is homeomorphic to the inverse image of $\{ x \}$ under the restriction of $pr_2$ to $C$ and is therefore compact [no. 2, Theorem 1 b)].

e) $\Longrightarrow$ a) : If $R$ is closed, then by definition $f$ is closed, and for each $z \in X/R$, $f^{-1}(z)$ is an equivalence class mod $R$ and is therefore compact; hence $f$ is proper by Theorem 1 b) of no. 2.

Finally we have to prove that $X/R$ is locally compact. $X'/R'$ is compact by c) and Proposition 8; the relation $R$ is that induced on $X$ by $R'$; $X$ is open in $X'$ and is saturated with respect to $R'$; hence $X/R$ is homeomorphic to the image $f'(X)$ of $X$ under the canonical mapping $f' : X' \to X'/R'$ ($§ 3$, no. 6, Proposition 10, Corollary 1). Now $f'(X)$ is open in $X'/R'$, and hence is a locally compact subspace of $X'/R'$.

Q.E.D.

f_z : X \to f(X) is proper [no. 1, Proposition 3 a)] it follows that X is compact (resp. locally compact) by reason of Proposition 6 of no. 2 and Proposition 7 of no. 3. Finally, if Y is compact (resp. locally compact) then so is f(X) which is closed in Y (no. 1, Proposition 1 and , no. 7, Proposition 13)

The first assertion is a consequence of the facts that each x \in X has a compact neighbourhood V and that f(V) is a compact neighbourhood of f(x) (§ 5, no. 3, Proposition 5 and § 9, no. 4, Theorem 2, Corollary 1). For each y \in K' let V(y) be a compact neighbourhood of some point of f^{-1}(y) in X, so that f(V(y)) is a compact neighbourhood of y. There are a finite number of points y_i \in K' such that the f(V(y_i)) cover K'. Let K_1 be the compact set $\bigcup_i V(y_i)$ in X; we have K' \subset f(K_1); hence K = K_1 \cap f^{-1}(K') is compact (because it is closed in K_1) and f(K) = K'.