A. Arhangel’skii and A. Taimanov
Unknown
Submitted 1960-01-01 | SovietRxiv: ru-196001.33021 | Translated from Russian

Abstract Generated abstract

This note strengthens a theorem of V. Ponomarev on extending open mappings to Čech compactifications by removing a compactness assumption in the single-valued case and giving a simpler proof in the multivalued setting. It proves that a continuous, closed and open mapping from a normal space onto a normal space has an open continuous extension from the Čech compactification of the domain onto that of the codomain. More generally, the same conclusion is established for strongly continuous, closed and open multivalued mappings satisfying the stated compactness condition on values. The proof uses a separation argument in normal spaces and a lemma excluding intersections between extended values and closures of complements of suitable images.

Full Text

A. Arhangel’skii and A. Taimanov

On a Theorem of V. Ponomarev

(Presented by Academician P. S. Aleksandrov, 11 VI 1960)

In paper (¹) V. Ponomarev proved for the first time that, under sufficiently broad assumptions, an open mapping \(f:X\to Y\) of a normal space \(X\) onto a normal space \(Y\) extends to an open mapping \(\varphi:\beta X\to \beta Y\) of the Čech extension \(\beta X\) onto the Čech extension \(\beta Y\). In doing so, V. Ponomarev obtained his result as a special case of a corresponding theorem for strongly continuous multivalued mappings. However, in V. Ponomarev’s theorem it was assumed that the mapping \(f\) is bicompact, i.e. that the inverse images \(f^{-1}y\) (and in the case of multivalued mappings also the images \(fx\)) of all points \(y\in Y\) (respectively \(x\in X\)) are bicompact.

One of the authors of the present note (A. Taimanov) freed (in the case of single-valued mappings) V. Ponomarev’s theorem from the condition of bicompactness. After this A. Arhangel’skii gave a considerably simpler proof of V. Ponomarev’s theorem strengthened in this way, extending it again to multivalued mappings.

Theorem 1. Let \(f\) be a continuous, simultaneously closed and open single-valued mapping of a normal space \(X\) onto a normal space \(Y\). Then the unique continuous extension \(\varphi:\beta X\to \beta Y\) of the mapping \(f\) is an open mapping of the space \(\beta X\) onto \(\beta Y\).

Theorem 2. Let \(f\) be a strongly continuous, simultaneously closed and open \(Y\)-bicompact mapping of a normal space \(X\) onto a normal space \(Y\). Then the unique strongly continuous extension* \(\varphi:\beta X\to \beta Y\) of the mapping \(f\) is an open mapping of the space \(\beta X\) onto \(\beta Y\).

Theorem 1 is contained in Theorem 2, to whose proof we now turn.

Notation. \(x\) is an arbitrary point of the space \(\beta X\); \(U\) is an arbitrary neighborhood of the point \(x\) in \(\beta X\); \(V\) is such a neighborhood of the point \(x\) in \(\beta X\) that \([V]\subset U\). All closures are denoted by square brackets and are considered in \(\beta X\), respectively in \(\beta Y\).

Lemma. The set \(\varphi x\cap [Y\setminus f(X\cap U)]\) is empty.

Suppose the contrary; then there exists a point \(y\in \varphi x\) contained in \([Y\setminus f(X\cap U)]\).

The sets
\[ B_0=f(X\cap [V])\subseteq f(X\cap U)\subseteq Y, \]
\[ B_1=Y\setminus f(X\cap U)\subseteq Y \]
are closed in \(Y\); the first because the mapping \(f\) is closed, the second because this mapping is open. Obviously, \(B_0\cap B_1\) is empty. Since \(Y\) is normal, it follows that \([B_0]\cap [B_1]\) is empty.

* Existing under our assumptions by virtue of the results of V. Ponomarev (²) (corollary of Theorem 2 on p. 270). In the present note we use the notation and terminology of papers (¹,²) (in particular, \(Y\)-bicompactness of the mapping \(f\) means that all \(f(x)\) are bicompact).

By our assumption, \(y \in [B_1]\). On the other hand, \(x \in [X \cap V] \subseteq [X \cap [V]]\), whence, by the strong and hence, in particular, oblique continuity of the mapping \(\varphi\), it follows that

\[ y \in \varphi x \subseteq [\varphi (X \cap [V])] = [f (X \cap [V])] = [B_0], \]

i.e. \(y \in [B_0] \cap [B_1]\). The contradiction obtained proves the lemma.

We prove Theorem 2, i.e. we prove that every point \(y \in \varphi x\) is an interior point of the set \(\varphi U\). Since \([V] \subset U\), it is enough to prove that \(y\) is an interior point of the set \(\varphi [V]\). By the lemma, \(y \notin [Y \setminus f(X \cap V)]\). The set \(\varphi [V] \subset \beta Y\) is bicompact; hence \(\beta Y \setminus \varphi [V]\) is open in \(\beta Y\). If \(y\) is not an interior point of the set \(\varphi [V]\), then in every neighborhood of it there are points of the set \(\beta Y \setminus \varphi [V]\) and, consequently, of the set \(Y \setminus \varphi [V]\). But then

\[ y \in [Y \setminus \varphi [V]] \subseteq [Y \setminus \varphi V] \subseteq [Y \setminus f(X \cap V)], \]

which contradicts the lemma. The theorem is proved.

Moscow State University
named after M. V. Lomonosov

Institute of Mathematics
Siberian Branch of the Academy of Sciences of the USSR

Received
11 V 1960

REFERENCES

  1. V. Ponomarev, DAN, 126, No. 4, 716 (1959).
  2. V. Ponomarev, DAN, 124, No. 2, 268 (1959).

Submission history

A. Arhangel’skii and A. Taimanov