UDC 513.831

MATHEMATICS

A. V. ARKHANGEL'SKII

A THEOREM ON THE METRIZABILITY OF THE PREIMAGE OF A METRIC SPACE UNDER AN OPEN-CLOSED FINITE-TO-ONE MAPPING, AN EXAMPLE, AND UNSOLVED PROBLEMS

(Presented by Academician P. S. Aleksandrov on 29 XII 1965)

It is proved that, under an open-closed finite-to-one mapping, the weight of the preimage does not exceed the weight of the image \((^2)\)*. This was preceded by a result of V. Proizvolov: if \(f: X \to Y\) is an open finite-to-one mapping of a locally bicompact space \(X\) onto a space \(Y\) of weight \(\leq \tau\), then the weight of \(X \leq \tau\) \((^3)\) (\(\tau\) is an infinite cardinal number). In connection with this, two problems arose:

1. Is every space that can be mapped openly, closedly, and finite-to-one onto a metric space metrizable?

2. Let \(f\) be an open finite-to-one mapping onto a compact space. Is there a countable base in the preimage (equivalently: is the preimage metrizable)?

Below a positive answer will be given to the first question and a negative one to the second. This progress is not a completion. On the contrary, only now do we come closely to V. Proizvolov’s problem:

1. Does every bicompactum which is mapped onto a compactum by an open mapping with compact preimages of points have a countable base?

2. In my opinion, the following question, concerning the relationship between feathered spaces and metric spaces, also deserves attention: does there exist in the first class a nonmetrizable space that is mapped openly and finite-to-one onto some space of the second class**?

The solution of the main problem (1) reduces to the proof of the following two assertions:

I. If \(f: X \to Y\) is an open finite-to-one mapping of an arbitrary space \(X\) onto a metric space \(Y\), then in \(X\) there is a sequence \(\varphi = \{\lambda_n\}\) of open covers \(\lambda_n\), satisfying the following quasi-refinement condition:

\[ \bigcap_{n=1}^{\infty} \lambda_n x = x \]

for every point \(x \in X\).

II. If a paracompact \(p\)-space has a quasi-refining sequence of open covers, then it is metrizable.

The theorem from \((^4)\), characterizing paracompact \(p\)-spaces as perfect preimages of metric spaces, now makes it possible to answer the first question.

Proof of I. We shall construct the sequence \(\{\lambda_n\}\) starting from the following standard representation of the mapping \(f: X \to Y\).

* In this paper only completely regular spaces and only continuous mappings are considered. The word “cover” denotes an ordinary open cover.

** A locally bicompact space that is mapped openly and finite-to-one onto a metric one is metrizable; this assertion is easily reduced to the above-mentioned result of V. V. Proizvolov.

Denote by \(Q_k\) the set of those points \(y \in Y\) whose inverse image \(f^{-1}y\) consists of exactly \(k\) points. Put \(P_k=f^{-1}Q_k\), \(Y_m=\bigcup_{k=1}^{m} Q_k\), \(X_m=\bigcup_{k=1}^{m} P_k(=f^{-1}Y_m)\), \(U_m=X\setminus X_m\), and \(V_m=Y\setminus Y_m\), \(k,m=1,2,\ldots,\ldots\infty\). Obviously, \(U_m=f^{-1}V_m\). Since \(f\) is open and finite-to-one, it follows that: a) all \(X_m,Y_m\) are closed; b) all mappings \(f_k=f|P_k\), \(f_k:P_k\to Q_k\), are closed; c) each \(f_k\) is a local homeomorphism. Condition c) means that each \(P_k\) is locally metrizable. But from the perfectness of the mappings \(f_k\) it follows that all \(P_k\) are paracompact. Hence all \(P_k\) are metrizable. Further, \(Y_m\) is a set of type \(G_\delta\) in \(Y\), and since \(X_m=f^{-1}Y_m\), \(U_m\) is the sum of a countable decreasing sequence of closed sets:

\[ U_m=\bigcup_{n=1}^{\infty} F_m^n . \]

In each of the spaces \(P_k\) choose a metrically decreasing sequence of open (in \(P_k\)) covers: \(\varphi_k=\{\gamma_k^n,\ n=1,2,\ldots,\ldots\infty\}\), \(k=1,2,\ldots,\ldots\infty\). For each \(G\in\gamma_k^n\), take some set \(\widetilde G\), open in \(X\), such that: 1) \(\widetilde G\cap P_k=G\); 2) \(\widetilde G\subseteq U_{k-1}\); 3) \(\widetilde G\cap F_k^n=\Lambda\).

Put \(\widetilde\gamma_k^n=\{\widetilde G\mid G\in\gamma_k^n\}\) and \(\lambda_n=\bigcup_{k=1}^{n}\widetilde\gamma_k^n\cup(U_n)\). We shall show that for any point \(x\in X\), \(\bigcap_{n=1}^{\infty}\lambda_n x=x\). Let \(x'\ne x\), \(x'\in X\). We have \(x\in P_{k_0}\subseteq U_{k_0-1}\), \(x'\in P_{k_1}\subseteq U_{k_1-1}\), for some \(k_0,k_1\). We want to prove that there is an \(n'\) such that \(\lambda_{n'}x\not\ni x'\). Let us note at once that this relation is equivalent to the following one: \(\lambda_{n'}x'\not\ni x\). We may therefore prove the first relation, assuming that \(k_0\le k_1\). For each \(k\), \(0<k\le k_0-1\), choose a number \(n(k)\) so that \(F_k^{n(k)}\ni x'\); this is possible, since \(\bigcup_{n=1}^{\infty}F_k^n=U_k\ni x'\) for \(0<k\le k_0-1\). Finally, choose \(n(k_0)\) from the condition \(\gamma_{k_0}^{\,n(k_0)}x\not\ni x'\), if \(x'\in P_{k_0}\), and from the condition \(x'\in F_{k_0}^{\,n(k_0)}\), if \(x'\in X\setminus P_{k_0}=U_{k_0}=\bigcup_{n=1}^{\infty}F_{k_0}^n\). Put \(N=\max_{1\le k\le k_0}(n(k)+k_0)\). Then

\[ \lambda_N x=\bigcup_{k=1}^{N}\widetilde\gamma_k^N x = \bigcup_{k=1}^{k_0-1}\widetilde\gamma_k^N x \cup \widetilde\gamma_{k_0}^N x . \tag{S} \]

By condition 3), no element of the system \(\gamma_k^N\), \(1\le k\le k_0-1\), intersects \(F_k^N\ni x'\). Therefore the first summand in the expression on the right of formula (S) does not contain the point \(x'\). If \(x'\notin P_{k_0}\), then the same argument shows that the second summand also does not contain the point \(x'\). If, however, \(x'\in P_{k_0}\), then the relation \(\widetilde\gamma_{k_0}^N x\not\ni x'\) follows from the equality \(\widetilde\gamma_{k_0}^N x\cap P_{k_0}=\gamma_{k_0}^N x\) (which follows from condition 1)) and from the fact that

\[ \gamma_{k_0}^N x\subseteq \gamma_{k_0}^{\,n(k_0)}x\subseteq X\setminus x' . \]

Consequently, \(\lambda_N x\not\ni x'\); since the points \(x,x'\) were chosen arbitrarily, we obtain \(\bigcap_{n=1}^{\infty}\lambda_n x=x\) for any point \(x\in X\). Assertion I is proved.

Proof of II. The following holds (see (4)).

Lemma. Let \(X\) be a topological space, and let \(\{\mu_n\}\) be a sequence of its open covers satisfying the conditions:

a) \(\bigcap_{n=1}^{\infty}\mu_n x=x\); b) for every \(x\in X\) and every \(n_1\) there is an \(n_2\) such that \([\lambda_{n_2}x]\subseteq \lambda_{n_1}x\), and for all \(n_2>n_1\) one has \(\lambda_{n_2}x\subseteq \lambda_{n_1}x\). Suppose, fur—

... \(bX\) is a bicompact Hausdorff extension of the space \(X\), and \(\{\eta_n\}\) is such a sequence of coverings of \(X\) by sets open in \(bX\) that: c)

\[ \bigcap_{n=1}^{\infty}\eta_n x \subseteq X; \]

d) \([\eta_{n_1}x]_{bX}\subseteq \eta_{n_2}x\) for \(n_1>n_2,\ x\in X\).

Then \(\{\xi_n\}\), where \(\xi_n=\eta_n\cap\mu_n,\ n=1,2,\ldots,\ldots\infty\), is a refining sequence of open coverings of the space \(X\).

To construct, for a paracompact \(p\)-space with a quasi-refining sequence of coverings \(\{\lambda_n\}\), sequences \(\{\mu_n\}\) and \(\{\eta_n\}\) with the properties a), b) and c), d), respectively, is not difficult. Put \(\mu_1=\lambda_1\), and suppose that \(\mu_n\) have already been defined for \(n\le k\). By \(\delta_k\) denote some covering of \(X\) by open sets each of whose closures is contained both in some element of \(\mu_k\) and in some element of \(\lambda_{k+1}\). As \(\mu_{k+1}\) take any locally finite covering inscribed in \(\delta_k\). The sequence \(\{\mu_k\}\) constructed by this rule satisfies conditions a) and b). Indeed, \(\mu_n x\subseteq \lambda_n x\) for every \(n\); therefore

\[ x\in\bigcap_{n=1}^{\infty}\mu_n x\subseteq\bigcap_{n=1}^{\infty}\lambda_n x=x, \]

i.e.

\[ \bigcap_{n=1}^{\infty}\mu_n x=x. \]

Further,

\[ [\mu_{n+1}x]=[\{\,\bigcup G\mid G\in\mu_{n+1},\ G\ni x\,\}] =\{\bigcup [G]\mid G\in\mu_{n+1},\ G\ni x\}\subseteq \mu_n x. \]

Now consider any bicompact Hausdorff extension \(bX\) of the space \(X\), and let \(\varphi=\{\gamma_n\}\) be some feathering of \(X\) in \(bX\), the elements of which form a decreasing sequence. We shall apply a procedure analogous to that carried out above.

Put \(\eta_1=\gamma_1\) and, assuming that \(\eta_k\) has already been defined, denote by \(\eta_{k+1}\) some covering of the space \(X\) by sets open in \(bX\) which cuts out on \(X\) a locally finite covering, the closures in \(bX\) of whose elements are contained both in elements of \(\eta_k\) and in elements of \(\gamma_k\). Continue the described process to infinity. Then

\[ \bigcap_{n=1}^{\infty}\eta_n x \subseteq \bigcap_{n=1}^{\infty}[\eta_n x\cap X]_{bX} = \bigcap_{n=1}^{\infty}[\eta_n\cap X]_{bX}x \subseteq \bigcap_{n=1}^{\infty}\gamma_n x \subseteq X^*. \]

Condition c) has been verified. Now

\[ [\eta_{n+1}x]_{bX}=[\eta_{n+1}\cap X]_{bX}x\subseteq\eta_n x. \]

Thus d) is also fulfilled. In view of the lemma formulated at the beginning and the theorem on metrizability of paracompacts with a refining sequence of coverings \((^{1,5})\), assertion II is proved. At the same time the following has been proved:

Theorem. The inverse image of a metric space under an open-and-closed finite-to-one mapping is metrizable.

Example. We shall construct a normal finally compact space that decomposes into the sum of two spaces with a countable base and that is mapped openly and finite-to-one onto an interval.

Let \(ABC\) be an equilateral triangle and let \(\{B_n\},\{C_n\}\) be such sequences of points lying respectively on the sides \(AB,AC\) that \(B_nC_n\parallel BC\), \(\rho(B_nC_n,BC)=1/n\). By \(X\) denote the set which is the union of the contours of the triangles \(ABC, AB_nC_n,\ n=1,2,\ldots,\ldots\infty\). Define the topology in \(X\) as follows: in \(X\setminus C\) it coincides with the topology induced by the surrounding plane; neighborhoods of the point \(C\) correspond to smooth curves \(S\), joining the point \(D\) of the half-interval \([AC)\) with the point \(C\), so that: 1) the set \(S\setminus(D\cup C)\) lies inside the triangle \(ABC\) and 2) \(S\) touches the line \(BC\) at the point \(C\)***. The neighborhood \(O_S C\) of the point \(C\) is that one of the two sets into which the triangle \(ABC\) is divided by the curve \(S\), which contains the interval \((DC)\), taken together with the point \(C\).

* By \([\eta_n\cap X]_{bX}\) is denoted the system formed by the closures in \(bX\) of the elements of the covering cut out on \(X\) by the system \(\eta_n\).

** In fact, we have just proved a more general assertion: a paracompact \(p\)-space that is mapped openly and finite-to-one onto a metric space is metrizable (see Question 4 and the example).

*** For brevity, in what follows we shall call such curves regular.

The space \(X\) has all the properties we need. To define an open finite-to-one mapping of it onto an interval presents no difficulty; it is induced, for example, by the orthogonal projection of \(X\) onto the interval \(AC\) (in the usual topology).

Obviously, \(X\) is the union of two spaces with a countable base:
\[ X=(X\setminus C)\cup(C). \]
The first is simply the product of a half-interval and an interval, and the second is a point. Therefore \(X\) is finally compact (there is a countable network in \(X\)). From the normality of \(X\setminus C\) it follows that, in order to prove the normality of \(X\), it suffices to establish the regularity of \(X\) at \(C\). The latter follows from the fact that for every proper curve \(S\) one can find a proper curve \(S'\) that lies wholly in \(O_S C\). Then
\[ [O_S C]=O_S C\cup S'\subset O_S C. \]

Remark 1. Well-known examples of finite-to-one mappings of nonmetrizable bicompacta onto compacta show that metrizability is not preserved when passing to the preimage, nor under closed finite-to-one mappings.

Remark 2. The most general formulation of a positive result in this circle of questions is now the following: if
\[ f:X\to Y \]
is a multivalued continuous open-closed mapping under which the preimages of points are compact, while the images of points are finite sets, then from the metrizability of \(X\) there follows the metrizability of \(Y\) (for the terminology, see \((^7)\)).

Indeed, \(f\) then decomposes into the superposition of two mappings\(^*\), the first of which is inverse to an open-closed finite-to-one single-valued mapping, and the second is a perfect single-valued mapping:
\[ X \xleftarrow{\,f_1\,} Z \xrightarrow{\,f_2\,} Y, \]
\[ \cap \]
\[ X\times Y \]
where
\[ Z=\{(x,y)\in X\times Y\mid y\in fx\}, \]
and \(f_1,f_2\) are induced by the projections of the product \(X\times Y\) onto its factors. One can now apply the theorem proved above (and conclude that if \(X\) is metrizable, then \(Z\) is metrizable) and Stone’s theorem \((^6)\) (and conclude that in this case \(Y\) is also metrizable).

Remark 3. The unsolved problem of V. V. Proizvolov (see \((^3)\)), the remark with which this paper begins, and the theorem proved above suggest the following hypothesis:

\((\Gamma)\). Metrizability is preserved under continuous open-closed multivalued mappings whose point-images are compacta (in particular, under mappings inverse to open perfect \(S\)-mappings).

Moscow State University
named after M. V. Lomonosov

Received
24 XII 1965

CITED LITERATURE

\(^1\) P. S. Aleksandrov, Bull. Polish Acad. Sci., Ser. Math., Astr. et Phys., 8, 135 (1960).
\(^2\) A. V. Arhangel’skii, DAN, 166, No. 6 (1966).
\(^3\) V. V. Proizvolov, DAN, 166, No. 1 (1966).
\(^4\) A. V. Arhangel’skii, Matem. sborn., 67 (109), No. 1, 55 (1965).
\(^5\) R. H. Bing, Canad. Math. J., 3, 175 (1951).
\(^6\) A. H. Stone, Proc. Am. Math. Soc., 7, 690 (1956).
\(^7\) V. I. Ponomarev, Matem. sborn., 51, No. 4, 515 (1960).

\(^*\) An arbitrary perfect multivalued mapping was represented in this form by Yu. M. Smirnov.