1
$\begingroup$

In [Tka] the author writes:

"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 [Eng]".

But I can not find this statement on this page. Is there a source in which this statement is explicitly formulated and proved?

[Tka] M. G. Tkachenko. Topological groups for topologists: part I, Bol. Soc. Mat. Mexicana (3), 5, 1999, 237-279.
[Eng] R. Engelking, General Topology, Heldermann Verlag, Berlin 1989.

$\endgroup$

1 Answer 1

1
$\begingroup$

It seems quite likely that it was intended to be page 331 of the Engelking's book published in 1977 in PWN, Warszava1 - rather then the later edition by Heldermann (revised and completed edition, 1989).

I will quote in full exercise 4.2.D (page 331 in the older edition, page 264 in the newer edition):

4.2.D. (a) (Ponomarev [1960]) Prove that a $T_0$-space $X$ is first countable if and only if $X$ is a continuous image of a metrizable space under an open mapping. Hint. Let $X$ be a first-countable space and let $\{U_s\}_{s\in S}$ be a base for $X$. Consider the Baire space $B(\mathfrak m) = \prod_{i=1}^\infty X_i$ where $X_i = S$ with the discrete topology, and the subset $T \subseteq B(\mathfrak m)$ consisting of all points $\{s_i\}$ such that $\{U_{s_i}\}_{i=1}^\infty$ is a base at a point $x\in X$; assign the point $x \in X$ to the point $\{s_i\}\in T$.
(b) (Michael [1971a]) Show that every first-countable space $X$ is a continuous image of a first-countable Hausdorff space under an open mapping, and deduce that in (a) the assumption that $X$ is a $T_0$-space can be omitted.
Hint (Shimrat [1956]). Define in $B(\mathfrak m)$, where $m = |X|$, a family $\{A_x\}_{x\in X}$ of pairwise disjoint dense subsets and consider the subset $\bigcup_{x\in X}(\{x\}\times A_x)$ of the Cartesian product $X \times B(\mathfrak m)$.
Remark. The construction in the above hint shows that every topological space is a continuous image of a Hausdorff space under an open mapping. Isbell proved more in [1969]: every topological space is a continuous image of a hereditarily paracompact and hereditarily strongly zero-dimensional space under an open mapping. Further information related to parts (a) and (b) can be found in Junnila [1978] and R. Pol [1981].
(c) (Arhangelskii [1963], Franklin [1965]) Observe that the construction outlined in Exercise 2.4.G(b) shows that sequential spaces can be characterized as the images of metrizable spaces under quotient mappings and Frechet spaces can be characterized as the images of metrizable spaces under hereditarily quotient mappings.

I will add that the same exercises is mentioned in the answer to this question on Mathematics Stack Exchange: Every Tychonoff space is an image of a Moscow space under a continuous open mapping. The references in that answer might be worth looking at, too.

1This older edition used to be available online - I do not know whether it is still accessible somewhere. Here is a Wayback Machine snapshot of the website where it used to be.

$\endgroup$
2
  • $\begingroup$ Although this seems to be very likely the place in Engelking's book that the author had in mind, I do not immediately see how this construction works for submetrizable spaces. (Hopefully, some people around here are familiar enough with submetrizable spaces to give some advice on that.) $\endgroup$ Oct 23, 2021 at 8:41
  • $\begingroup$ The answer is contained in Junnila's paper. From the scanned page to which you have: "every topological space is the continuous image of a sigma-discrete stratifiable space under an open map" (every paracompact space with a $G_\delta$-diagonal is submetrisable). $\endgroup$
    – Tyrone
    Oct 25, 2021 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.