Theorem 1

*Let $f : X \to Y$ be a continuous mapping. Then the following four statements are equivalent:

a) $f$ is proper. b) $f$ is closed and $\overline{f}^{-1}(y)$ is quasi-compact for each $y \in Y$. c) If $\mathfrak{F}$ is a filter on $X$ and if $y \in Y$ is a cluster point of $f(\mathfrak{F})$ then there is a cluster point $x$ of $\mathfrak{F}$ such that $f(x) = y$. d) If $\mathfrak{U}$ is an ultrafilter on $X$ and if $y \in Y$ is a limit point of the ultrafilter base $f(\mathfrak{U})$, then there is a limit point $x$ of $\mathfrak{U}$ such that $f(x) = y$.

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: