Every Tychonoff space is an image of a Moscow space under a continuous open mapping.

Every Tychonoff space is an image of a Moscow space under a continuous open mapping.

A space $X$ is called Moscow if the closure of every open set $U\subset X$ is the union of a family of $G_\delta$-subsets of $X$.

Reference: A. Arhangel'skii and M. Tkachenko, Topological Groups and Related Structures, p358, Ex 6.3.b.

I've tried to answer the question but I did not succeed. For example, For Tychonoff space $X$, if we can construct a space $Y$ such that $Y\times X$ is extremally disconnected (Or Moscow) then $\pi:X\times Y\to X$ is continuous open mapping but it is impossible.

This is a story.

I did not read the book by Arhangel'skii and Tkachenko, I only glanced at some pages of it related with my investigations. Since I found that even some of “open problems” related to my investigations are easily solvable or already solved, I decided that the exercises should not be very hard. :-) But I am not a specialist in Moscow spaces. :-( Even more, maybe I firstly read about them from the same book. :-) So, maybe there is a trick in the solution of this exercise.

Searching the answer to the you question, I looked in Bible of general topologists – “General topology” [Eng] by Ryszard Engelking, and found unexpectedly few helpful claims. In fact, there was only a remark in the Exercise 4.2.D in my (Russian) edition, claiming that Isbell in [Isb] proved that each topological space is an open image of a hereditarily paracompact and strongly zero-dimensional space. But we need somewhat stronger result, because, by Theorem 6.2.25 from [Eng], we have that each nonempty extremally disconnected Tychonoff space in strongly zero-dimensional, but the opposite inclusion is not true (the space of rational numbers is strongly zero-dimensional but not extremally disconnected).

So I tried to look at Isbell’s paper [Isb]. And I successfully found it here, at Springer. But this day was not mine, :-) because article’s free preview ends exactly at the searched theorem, :-) and idea of paying for such the things is completely alien to my mentality. :-)

So I had to find other ways. Unexpectedly, I found few results in the Internet too.

According to [Haz, p.524, Tychonoff spaces are perfect images of extremally disconnected spaces. But this should not work, because, accordingly to [Eng], an open image of an of extremally disconnected space is extremally disconnected too.

From the other side, it seems that the original problem can be reduced to compact spaces. Indeed, let $X$ be a Tychonoff space. There exists a compact space $bX$ such that $X$ is a dense subset of $bX$. If $f:Y\to bX$ is a continuous open map from a Moscow space $Y$ onto $bX$, then $f^{-1}(X)$ is dense in $Y$. Since a dense suspace of a Moscow space is Moscow [AT, Prop. 6.1.2], we see that $f^{-1}(X)$ is a Moscow space. But, by the above remark, $f^{-1}(X)$ cannot be extremally disconnected. In particular, Martin’s construction of the map from $\beta X$ showing that Every compact space is a continuous image of a compact Moscow space. fails for the present proof. Moreover, if $f$ is an open continuous map of a Moscow space $Y$ onto a space $X$ such that $f^{-1}(x)$ is compact for each $x\in X,$ then $X$ is also a Moscow space [AT, Th. 6.3.1]. So, maybe for the construction of the desired space $Y$ we should “spread vertically” the space $X$ in the product $X\times Z$ for some space $Z$.

At last I got it. I found the theorem which should imply the positive answer.

A (Hausdorff) space is called strictly $\sigma$-discrete, if it is a union of countably many of its closed discrete subspaces. It is easy to show that each point of a strictly $\sigma$-discrete space is a $G_\delta$-set. Therefore each strictly $\sigma$-discrete space is Moscow. According to [Tka], Junilla (who also was mentioned in the Engelking’s remark) showed in [Jun] that every topological space is an open image of a strictly $\sigma$-discrete Tychonoff space. $\square$

Update. I have just found an other reference in [Tka’]: “Every topological space $X$ can be represented as an open continuous image of a completely regular submetrizable space $Y$ (in other words, $Y$ admits a continuous one-to-one mapping onto a metrizable space) – the corresponding construction is given on p. 331 of [Eng2]”.

References

[AT] A.V. Arhangel'skii, M. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

[Eng] R. Engelking. General Topology. -- M.: Mir, 1986 (in Russian).

[Eng2] R. Engelking. General Topology. -- Heldermann Verlag, Berlin, 1989.

[Haz] M. Hazewinkel (ed.), Encyclopaedia of Mathematics (9).

[Isb] J.R. Isbell. A note on complete closure algebras. -- Math. Systems. Theory 3 (1969), 310-312.

[Jun] H.J.K. Junnila, Stratifiable preimages of topological spaces, Colloq. Math. Soc. J. Bolyai 23. Topology, Budapest, 1978, 689--703. MR **81m:*54019

[Tka’] M. G. Tkachenko. Topological groups for topologists: part I, Bol. Soc. Mat. Mexicana (3), 5, 1999, 237-279.

[Tka] V.V. Tkachuk. When do connected spaces have nice connected preimages? -- Proc. Amer. Math. Soc, 125:11 (November 1998), 3437--3446.