V. V. Proizvolov

On One-to-One Mappings onto Metric Spaces

(Presented by Academician P. S. Aleksandrov on 24 IV 1964)

The general question answered by the theorems of this paper is the following: when, under a condensation,* is the preimage of a metric space metrizable?

Theorem 1. Let a condensation \(f:X\to Y\) be given, where \(X\) is a locally connected and peripherally bicompact space, and \(Y\) is a metric space. Then \(X\) is metrizable.**

The space \(Y\), by Bing’s metrization criterion \((^{1})\), has a \(\sigma\)-discrete base \(\gamma=\{\gamma_i\}\), where each \(\gamma_i\) is a discrete system of open sets:
\[ \gamma_i=\{U_{i\alpha}\},\qquad U_{i\alpha}\cap U_{i\beta}=\Lambda,\quad \alpha\ne\beta;\quad i=1,2,\ldots \]

Denote by \(V_{i\alpha}\) the \(1/i\)-neighborhood of the set \(U_{i\alpha}\), i.e. the set of all points whose distance from \(U_{i\alpha}\) is not greater than \(1/i\). The collection of all \(V_{i\alpha}\) forms a base of the space \(Y\); denote it by \(\omega=\{V_{i\alpha}\}\).

Let us note that every open subset of a locally connected space decomposes into open components. Consequently,
\[ f^{-1}V_{i\alpha}=\{W_{i\alpha\delta}\}, \]
where each \(W_{i\alpha\delta}\) is open and connected,
\[ W_{i\alpha\delta_1}\cap W_{i\alpha\delta_2}=\Lambda,\quad \delta_1\ne\delta_2. \]
The union of all \(W_{i\alpha\delta}\) over all indices \(i,\alpha,\delta\) is a system of open subsets of the space \(X\); denote it by \(W\). We shall verify that \(W\) is a base in \(X\).

Take an arbitrary point \(x\) with its neighborhood \(Ox\). We must show that there is a \(W_{i\alpha\delta}\) such that
\[ x\in W_{i\alpha\delta}\subset Ox. \]
Take a connected neighborhood \(O'x\subset Ox\) with bicompact boundary. Denote
\[ F=[O'x]\setminus O'x. \]
The set \(F\) is bicompact, and therefore \(fF\) is also bicompact, not containing the point
\[ y=fx. \]
There is an element \(V_{i\alpha}\in\omega\) such that
\[ y\in V_{i\alpha},\qquad V_{i\alpha}\cap fF=\Lambda. \]
Take that component of the set \(f^{-1}V_{i\alpha}\) which contains \(x\); let it be \(W_{i\alpha\delta_0}\). It is asserted that
\[ W_{i\alpha\delta_0}\subset O'x. \]
If there were a point
\[ a\in W_{i\alpha\delta_0}\setminus O'x, \]
then the set \(W_{i\alpha\delta_0}\), being connected, would intersect \(F\), but then
\[ V_{i\alpha}\cap fF\ne\Lambda, \]
which is not the case. Thus,
\[ W=\{W_{i\alpha\delta}\} \]
is a base of the space \(X\).

Denote
\[ Z_{i\alpha\delta}=W_{i\alpha\delta}\cap f^{-1}U_{i\alpha}. \]
It is asserted that
\[ Z=\{Z_{i\alpha\delta}\} \]
is a base in \(X\). Take an arbitrary point \(x\) with its neighborhood \(Ox\). We must show that there is a \(Z_{i\alpha\delta}\) such that
\[ x\in Z_{i\alpha\delta}\subset Ox. \]
Take a connected neighborhood \(O'x\subset Ox\) with bicompact boundary. Denote
\[ F=[O'x]\setminus O'x. \]
The set \(F\) is bicompact, and therefore \(fF\) is also bicompact. There is an element \(U_{i\alpha}\in\gamma\) such that
\[ y\in U_{i\alpha} \quad\text{and}\quad V_{i\alpha}\cap fF=\Lambda, \]
where
\[ y=fx. \]
Take that component of the set \(f^{-1}V_{i\alpha}\) which contains \(x\); let it be \(W_{i\alpha\delta_0}\). It is easy to verify that
\[ W_{i\alpha\delta_0}\subset O'x, \]
and consequently
\[ Z_{i\alpha\delta_0}\subset O'x, \]
and, since
\[ x\in Z_{i\alpha\delta_0}, \]
it has been proved that
\[ Z=\{Z_{i\alpha\delta}\} \]
is a base of the space \(X\).

It remains to prove that
\[ Z=\{Z_{i\alpha\delta}\} \]
is a \(\sigma\)-discrete system of sets of the space \(X\). Fix \(i\) and consider
\[ Z_i=\{Z_{i\alpha\delta}\}; \]
we shall prove that this is a discrete system of sets in \(X\). First, this system

* A condensation is a one-to-one continuous mapping.
* A space is peripherally bicompact if it has a base whose elements have bicompact boundaries.
** Theorem 1 overlaps with Mardesic’s theorem that a locally connected bicompactum mapped zero-dimensionally onto a compactum is metrizable.

disjoint. Suppose that it is not discrete, i.e., suppose that there is a point \(x\in X\) such that every neighborhood \(Ox\) of it intersects more than one set of the system \(Z_i=\{Z_{i\alpha\delta}\}\). Then, obviously, there is an \(\alpha\) such that \(y\in [U_{i\alpha}]\), where \(y=fx\). Take an arbitrary neighborhood \(Oy\) of the point \(y\), but such that \(Oy\subset V_{i\alpha}\), and such that \(Oy\) intersects only one element of the system \(\gamma_i\), namely \(U_{i\alpha}\).

Now take that component of the set \(f^{-1}Oy\) which contains the point \(x\), and denote it by \(Ox\). There is a \(\delta_0\) such that \(Ox\subset W_{i\alpha\delta_0}\), and, therefore, the neighborhood \(Ox\) intersects only one element of the system \(Z_i=\{Z_{i\alpha\delta}\}\), namely \(Z_{i\alpha\delta_0}\). This contradicts the original assumption. Thus, \(Z=\{Z_{i\alpha\delta}\}\) is a \(\sigma\)-discrete base in \(X\), and, by Bing’s criterion, the space \(X\) is metrizable.

Theorem 2. Suppose a condensation \(f:X\to Y\) is given, where \(X\) is a locally connected, connected, and locally bicompact space, and \(Y\) is a Hausdorff, locally bicompact, locally connected, unicohesive space with a countable base. Then \(f\) is a homeomorphism.

Corollary. Suppose a condensation \(f:X\to E^n\) is given, where \(X\) is a locally connected, connected, and locally bicompact space, and \(E^n\) is Euclidean space. Then \(f\) is a homeomorphism.

Theorem 2 follows from considering Theorem 1 together with the main theorem of the paper \((^2)\).

A. V. Arkhangel’skii constructed an example of a condensation of a locally bicompact nonmetrizable space onto a metric space. This example shows that in Theorem 1 the condition of local connectedness of the space \(X\) is essential.

Theorem 3. Suppose there is a condensation \(f:X\to Y\), where \(X\) is a countably compact\(^*\) space and \(Y\) is a metric space. Then \(f\) is a homeomorphism, and \(X\) is a compactum\(^ {**}\).

Proof. The continuous image of a countably compact space is countably compact. Consequently, \(Y\) is a compactum. It remains only to prove that \(f\) is a homeomorphism. If a condensation is a closed mapping, then it is a homeomorphism. Suppose that \(f\) is not closed, i.e., that there is a closed set \(A\subset X\) such that \(B=fA\) is not closed in \(Y\). In the compactum \(Y\) there is a countable subset \(C\subset B\) whose unique limit point, the point \(a\), lies outside \(B\). Thus, \(C=\{c_i\}\); for each point \(c_i\) choose a neighborhood \(Oc_i\) such that \([C]\cap Oc_i=c_i\).

Now denote \(U=Y\setminus [C]\). Next, take a neighborhood \(Ob\) of the point \(b=f^{-1}a\) such that \(Ob\cap f^{-1}C=\Lambda\).

I assert that from the countable covering of the space \(X\), consisting of \(f^{-1}U\), \(Ob\), and all \(f^{-1}Oc_i\), \(i=1,2,\ldots\), it is impossible to choose a finite covering; this contradicts the countable compactness of the space \(X\). The theorem is proved.

It follows from Theorem 3, for example, that the space of ordinal numbers up to \(\omega_1\) exclusively cannot be condensed onto a metric space.

In connection with Theorem 2 I pose a question, the answer to which is unknown to me. Is it possible to condense onto \(E^n\) a connected locally bicompact space of uncountable weight?

Moscow State University
named after M. V. Lomonosov

Received
17 IV 1964

REFERENCES

1. R. H. Bing, Canad. J. Math., 3, No. 2, 175 (1951).
2. V. Proizvolov, DAN, 151, No. 6, 1286 (1963).

* A space is countably compact if from every countable covering of it one can choose a finite subcovering.

** Note added in proof. As has become known to me, a similar result, as well as some other results in this direction, were independently obtained by V. Klyushin.