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Theorem: Assume that $X$ is Čech-complete, and $Y$ is paracompact. If there exists $f\colon X\to Y$ which is surjective, open and continuous then $Y$ is Čech-complete.

The theorem appears as a corollary in a paper by Michael [1] where he proves a much stronger theorem. In the same paper he refers to papers by Pasynkov [2] and by Worrell [3] proving (independently) the above theorem.

I could not find either papers online or in my local library. Since I haven't got much experience with Čech-complete spaces, nor with paracompact spaces (I mean of course general paracompact spaces, I do have some experience with metric spaces) I am ignorant of "standard tricks", and was wondering if anyone can give me a sketch of the proof, or help me locate an online (or different than the three cited here) reference for the proof.


  1. Michael, E. Complete spaces and tri-quotient maps. Illinois J. Math. 21, no. 3 (1977), pp. 716–733.

  2. Pasynkov, B. Open mappings. Dokl. Akad. Nauk USSR, vol. 175 (1967), pp. 292-295; Soviet Math. Dokl., vol. 8 (1967), pp. 853-856.

  3. Worrell, J.M., Jr. The paracompact open continuous images of Čech-complete spaces, Notices Amer. Math. Soc., vol. 13 (1966), p. 858.

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I couldn't resist... – t.b. Jun 1 '12 at 7:32
@t.b.: Yes, I knew it was you. It was too obvious! :-) I was hoping someone would do it, and I had a hunch it was you when I saw the downvote... Thanks! :-) – Asaf Karagila Jun 1 '12 at 7:33
I didn't even try to hide :) Have fun with the badges and the screen shot :) – t.b. Jun 1 '12 at 7:44
up vote 3 down vote accepted

Engelking has it as part of an exercise (5.5.8) with hints; I’ve added some details.

Suppose that $X$ is Čech-complete, $Y$ is paracompact, and $f:X\to Y$ is a continuous open surjection. We show first that there is a closed $G_\delta$-set $A \subseteq X$ such that $f\upharpoonright A$ is a perfect map of $A$ onto $Y$. (A perfect map is a continuous closed map with compact fibres.) To do this, let $F:\beta X \to \beta Y$ be the Čech-Stone extension of $f$. Let $Z = F^{-1}[Y]$, and let $g = F\upharpoonright Z$; clearly $g$ is perfect. Suppose that $G$ is an open subset of $Z$ such that $f[G \cap X] = Y$.

Claim: There is an open subset $H$ of $Z$ such that $f[H \cap X] = Y$ and $\operatorname{cl}_ZH \subseteq G$.

Proof of Claim: For each $y \in Y$ let $V(y)$ be an open set in $Z$ such that $f^{-1}[\{y\}] \cap V(y) \ne \varnothing$ and $\operatorname{cl}_ZV(y) \subseteq G$. Let $\mathscr{U}$ be a locally finite open refinement of the open cover $\{f[V(y) \cap X]:y \in Y\}$ of the paracompact space $Y$. For each $U \in \mathscr{U}$ choose a point $y_U \in Y$ such that $U \subseteq f[V(y_U) \cap X]$, and let $$H = \bigcup\limits_{U \in \mathscr{U}} \left(V(y_U) \cap g^{-1}[U]\right).$$ Clearly $H$ is open in $Z$. Given $y \in Y$, fix $U \in \mathscr{U}$ such that $y \in U$. Then $y \in f[V(y_U) \cap X]$, so we can choose $x \in V(y_U) \cap X$ such that $y = f(x)$. But then $g(x) = F(x) = f(x) = y \in U$, so $x \in V(y_U) \cap g^{-1}[U] \cap X \subseteq H \cap X$, and $y \in f[H \cap X]$. It only remains to show that $\operatorname{cl}_ZH \subseteq G$.

For each $U \in \mathscr{U}$ let $W_U = V(y_U) \cap g^{-1}[U]$, and let $\mathscr{W} = \{W_U:U \in \mathscr{U}\}$, so that $H = \bigcup \mathscr{W}$. For each $U \in \mathscr{U}$ we have $\operatorname{cl}_ZW_U \subseteq \operatorname{cl}_ZV(y_U) \subseteq G$. Suppose that $z \in Z \setminus \bigcup\limits_{U \in \mathscr{U}}\operatorname{cl}_ZW_U$. $\mathscr{U}$ is locally finite, so $g(z)$ has an open nbhd $N$ that meets only finitely many members of $\mathscr{U}$. Then $g^{-1}[N]$ is a $Z$-open nbhd of $z$ that meets only finitely many members of $\mathscr{W}$. $Z$ is regular, so $g^{-1}[N]$ can be shrunk to a $Z$-open nbhd of $z$ disjoint from $\bigcup\limits_{U\in\mathscr{U}}\operatorname{cl}_ZW_U \supseteq H$, and hence $z \notin \operatorname{cl}_ZH$. Thus, $\operatorname{cl}_ZH = \bigcup\limits_{U\in \mathscr{U}}\operatorname{cl}_ZW_U \subseteq$ $G$. $\dashv$

Now because $X$ is $X$ is Čech-complete, it’s a $G_\delta$ in $\beta X$ and hence in $Z$, so we can write $X = \bigcap\limits_{n\in\omega}G_n$, where each $G_n$ is open in $Z$. Let $H_0 = Z$, and for $n\in\omega$ let $H_{n+1}$ be an open subset of $Z$ such that $\operatorname{cl}_ZH_{n+1} \subseteq G_n \cap H_n$ and $f[H_{n+1} \cap X] = Y$; the Claim ensures that this is possible. Let $A = \bigcap\limits_{n\in\omega}H_n = \bigcap\limits_{n\in\omega}\operatorname{cl}_ZH_n$; clearly $A$ is a closed $G_\delta$-set in $X$. $A$ is also closed in $Z$, so $f\upharpoonright A = g\upharpoonright A$ is perfect. Finally, suppose that $y \in Y$. For each $n \in \omega$ let $K_n = \operatorname{cl}_ZH_n \cap X \cap f^{-1}[\{y\}] \ne \varnothing$. Then $\{K_n:n \in \omega\}$ is a decreasing nest of closed subsets of the compact set $f^{-1}[\{y\}]$, so its intersection is a non-empty subset of $A$, and hence $y \in f[A]$.

Now $\operatorname{cl}_{\beta X}A = \beta A$, and $f_0 \triangleq f\upharpoonright A$ is perfect, so $F_0 \triangleq F\upharpoonright \operatorname{cl}_{\beta X}A$ satisfies the condition $$F_0[\operatorname{cl}_{\beta X}A\setminus A] \subseteq \beta Y \setminus Y.\tag{1}$$ This is a standard result, so I’ll just sketch the argument. Let $Z_0 = F_0^{-1}[Y]$; then $F_0\upharpoonright Z_0$ is a continuous extension of $f_0$ with the same range, so $f_0 = F_0 \circ id_A$. Since the composition $F_0 \circ id_A$ is perfect, $id_A:A\to Z_0$ must be perfect. (This takes a bit of argument, which I can supply if necessary.) It follows that $A = id_A[A]$ is a closed subset of $Z_0$. But $A$ is dense in $\operatorname{cl}_{\beta X}A \supseteq Z_0$, so we must have $Z_0 = A$, which immediately implies $(1)$.

$A$ is a closed subset of the Čech-complete space $X$, so $A$ is Čech-complete and is therefore a $G_\delta$-set in $\beta A = \operatorname{cl}_{\beta X}A$. Of course $\beta A \setminus A$ is an $F_\sigma$-set in $\beta A$, so we can write $\beta A \setminus A = \bigcup\limits_{n \in \omega}C_n$, where each $C_n$ is closed in $\beta A$. $F_0$ is perfect and hence closed, so the sets $F_0[C_n]$ are closed in $\beta Y$, and $(1)$ ensures that $F_0[C_n] \subseteq \beta Y \setminus Y$ for each $n \in \omega$. $F_0[A] = Y$, so in fact $\beta A \setminus A = \bigcup\limits_{n\in\omega}F_0[C_n]$ is an $F_\sigma$-set in $\beta Y$, and hence $Y$ is Čech-complete.

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