UDC 519.50 + 519.54

MATHEMATICS

P. K. OSMATESKU

ON THE EXTENSION OF PERFECT MAPPINGS TO \(\omega\alpha\)-EXTENSIONS

(Presented by Academician P. S. Aleksandrov, 19 III 1965)

E. G. Sklyarenko in the paper \((^1)\) proved the following theorem:

Let \(\bar X\) and \(\bar Y\) be bicompact extensions of completely regular spaces \(X\) and \(Y\), with \(\bar X\) perfect and \(\bar Y\) having a punctiform remainder. Every perfect mapping \(f: X \to Y\) extends to a mapping

\[ \bar f:\bar X \to \bar Y . \]

In the present paper this result is generalized to the case of \(T_1\)-spaces and \(\omega\alpha\)-extensions.

Lemma 1. Let two \(T_1\)-spaces \(X\) and \(Y\) and a perfect mapping \(f:X\to Y\) be given. Then the mapping \(f\) extends to a perfect mapping \(\bar f\) of the space \(\omega X\) onto \(\omega Y\), taking the remainder \(\omega X\setminus X\) into the remainder \(\omega Y\setminus Y\).

Proof. By Theorem 1.3 of A. V. Arkhangel’skii \((^{2,3})\), the mapping \(f\) extends (uniquely) to a continuous closed mapping \(\bar f\) of the space \(\omega X\) onto \(\omega Y\). The mapping \(\bar f\) takes the remainder \(\omega X\setminus X\) into the remainder \(\omega Y\setminus Y\). Indeed, suppose that for some point \(\eta\in Y\) the preimage \(f^{-1}\eta\) contains at least one point \(\xi\in \omega X\setminus X\). The set \(f^{-1}\eta\subset X\) is closed in \(X\) and, by the condition of the lemma, bicompact. Since \(\omega X\) is an extension, \([f^{-1}\eta]_{\omega X}=f^{-1}\eta\) (Definition 2 \((^4)\)). Consequently, the point \(\xi\) cannot be a limit point for \(f^{-1}\eta\). But it also cannot be isolated, because, by the construction of the mapping \(\bar f\) (Theorem 1 \((^2)\)), the point \(\xi\) must have as its coordinate the set \(f^{-1}\eta\); in view of these circumstances the supposition that the preimage \(\bar f^{-1}\eta\) contains points from \(\omega X\setminus X\) is false. The lemma is proved.

Remark 1. In what follows, without making a special stipulation, we shall denote a point of the space of an upper semicontinuous decomposition by the same symbol as the body of the corresponding element of the decomposition, only in parentheses; for example, \((A)\) is the point of the space of the upper semicontinuous decomposition corresponding to the element of the decomposition whose body is the set \(A\).

Remark 2. Without making a special stipulation, by \(\omega\alpha X\) we shall denote the \(\omega\alpha\)-extension of the space \(X\) (Definition 2 \((^4)\)).

Lemma 2. For any centered and relatively maximal with respect to centeredness system \(\gamma=\{P_\beta\}\) of closed sets of the space \(X\), the set \(\bigcap_\beta [P_\beta]_{\omega\alpha X}\) consists of one point.

Proof. Let \(\gamma=\{P_\beta\}\) be an arbitrary maximal centered system of closed sets of the space \(X\). We shall prove that \(\bigcap_\beta [P_\beta]_{\omega\alpha X}\) consists of one point.

First of all,

\[ \bigcap_\beta [P_\beta]_{\omega\alpha X}\ne \Lambda \]

in view of the closedness of \([P_\beta]_{\omega\alpha X}\) and the bicompactness of \(\omega\alpha X\). The sets of the system \(\gamma\) are coordinates of some point of the space \(\omega X\): \(\xi=\{P_\beta\}\). We shall show that some point \((A)\in \omega\alpha X\) is a point of contact of the system \(\gamma\) in that

* A continuous mapping is called perfect if it is closed and the full preimages of all points are bicompact.

if and only if \(\xi \in A\). In one direction the assertion is obvious, i.e., if \(\xi \in A\), then \((A) \in \bigcap_\beta [P_\beta]_{\omega\alpha X}\). Conversely, if \(\xi \in A\), then \((A) \in \bigcap_\beta [P_\beta]_{\omega\alpha X}\). Indeed, if \(\xi \in A\), then for an arbitrary point \(\xi_A \in A\) there is a set \(P_{\beta_1} \in \xi\) such that \(\xi_A \in [P_{\beta_1}]_{\omega X}\); hence the point \(\xi_A\) has a neighborhood \(O\xi_A\) not meeting the set \(P_{\beta_1}\). The neighborhoods \(\{O\xi_A\}\) chosen in this way form a cover of the set \(A\), from which, by the bicompactness of \(A\), one can choose a finite cover \(\{O\xi_{A_i}\}\). Put \(OA=\bigcup_{i=1}^k O\xi_{A_i}\supset A\); \(P_A=\bigcap_{i=1}^k P_{\beta_i}\), where from \(P_{\beta_i}\in\xi\) it follows that \(P_A\in\xi\). Since \(P_{\beta_i}\cap O\xi_{A_i}=\Lambda\), we have \(P_A\cap OA=\Lambda\); hence \((A)\in [P_A]_{\omega\alpha X}\), and consequently \((A)\in \bigcap_\beta [P_\beta]_{\omega\alpha X}\). But there exists one and only one element of the partition containing \(\xi\); denote it by \(A_\xi\). Hence, by what was proved above, it follows that
\[ (A_\xi)=\bigcap_\beta [P_\beta]_{\omega\alpha X}. \]

Remark. For all points \(\xi=\{P_\beta\}\) we have
\[ \bigcap_\beta [P_\beta]_{\omega\alpha X}=(A). \]

Lemma 3. Let \(f\) be a continuous mapping of the space \(\omega\alpha X\) onto \(\omega\alpha Y\). If the mapping \(f\) on \(X\) is perfect and \(fX=Y\), then the relation
\[ f(\omega\alpha X\setminus X)=\omega\alpha Y\setminus Y \]
holds.

Proof. The mapping \(f:\omega\alpha X\xrightarrow{\text{onto}}\omega\alpha Y\) generates a certain mapping \(g=f\varphi\) of the space \(\omega X\) onto \(\omega\alpha Y\), where \(\varphi\) is the natural mapping of the space \(\omega X\) onto \(\omega\alpha X\), which exists by Definition 2 \((^4)\); the mapping \(f\) induces a certain mapping \(f_0\) on \(X\), which, by the hypothesis of the lemma, is perfect. On the basis of Lemma 1, \(f_0\) extends (uniquely) to a perfect mapping \(\bar f_0\) of the space \(\omega X\) onto \(\omega Y\), carrying the remainder \(\omega X\setminus X\) into the remainder \(\omega Y\setminus Y\).

The mapping \(\bar f_0\) generates a certain continuous mapping \(g_0\) of the space \(\omega X\) onto \(\omega\alpha Y\), which is defined as follows:
\[ g_0\xi=\bar f_0\xi=f_0\xi,\qquad \text{if } \xi\in X; \]
\[ g_0\xi=(B),\qquad \text{if } \bar f_0\xi\in B \text{ or } g_0=\psi\bar f_0, \]
where \(\psi\) is the natural mapping of the space \(\omega Y\) onto \(\omega\alpha Y\), which exists by Definition 2 \((^4)\). We obtain the diagram:
\[ \begin{array}{ccc} \omega X & \xrightarrow{\ \bar f_0\ } & \omega Y\\ \varphi \downarrow & & \downarrow \psi\\ \omega\alpha X & \xrightarrow{\ f\ } & \omega\alpha Y \end{array} \]

Let us prove that this diagram is commutative, i.e., that the mapping \(g_0=\psi\bar f_0\) coincides with the mapping \(g=f\varphi\). At the points of the space \(X\) this is obvious. Let \(\xi\in\omega X\setminus X\). The maximal centered system of closed sets \(\xi=\{P_\alpha\}\) of the space \(X\) is carried, under the mapping \(f_0\), into a maximal centered system of closed sets \(\eta=\{f_0P_\alpha\}\) of the space \(Y\). The centeredness of the system \(\eta\) is obvious. We shall show its maximality. Suppose the contrary, i.e., suppose that the centered system \(\eta\) is not maximal; then there exists such a ...

a closed set \(F \subset Y,\ F \in \eta\), upon adding which to the system \(\eta\) we again obtain a centered system. Hence it follows that for an arbitrary finite subsystem of closed sets \(P_{\alpha_1}, P_{\alpha_2}, \ldots, P_{\alpha_k}\) of the system \(\xi=\{P_\alpha\}\) we have
\[ f_0P_{\alpha_1}\cap f_0P_{\alpha_2}\cap \cdots \cap f_0P_{\alpha_k}\cap F\ne \Lambda \]
or else
\[ P_{\alpha_1}\cap P_{\alpha_2}\cap \cdots \cap P_{\alpha_k}\cap f_0^{-1}F\ne \Lambda, \]
where \(f_0^{-1}F\in \xi\). Thus the system \(\xi\), together with the set \(f_0^{-1}F\), again forms a centered system, contrary to the maximality of \(\xi\).

The contradiction obtained proves the maximality of the system \(\eta\). By Theorem 2 \((^4)\), \(\omega\alpha Y\) is the space of some continuous external decomposition of the space \(\omega Y\); hence the point \(\eta\in\omega Y\) belongs to some class \(B\subset \omega Y\). By the definition of \(f_0\) (see Theorem 1 \((^2)\)), \(\bar f_0\xi=\{f_0P_\alpha\}\); consequently, \(\{f_0P_\alpha\}=\bar f_0\xi\in B\), and hence, by the definition of \(g_0\), we have \(g_0\xi=(B)\). Let \(g\xi=(B')\); we shall prove that \((B')=(B)\), thereby proving the coincidence of the mappings \(g\) and \(g_0\).

As we saw above, \(\bar f_0\xi=\{f_0P_\alpha\}\in B\); hence, on the basis of Lemma 2 \((^4)\),
\[ \bigcap_\alpha [f_0P_\alpha]_{\omega\alpha Y}=(B). \]
But \(\xi=\bigcap_\alpha [P_\alpha]_{\omega X}\). By Lemma 1 \((^4)\), the mapping \(g\) is closed, and moreover
\[ g[P_\alpha]_{\omega X}=[gP_\alpha]_{\omega\alpha Y}=[f_0P_\alpha]_{\omega\alpha Y} \]
(since \(g\) coincides with \(f_0\) on \(X\)); consequently,
\[ g\xi=(B')\in \bigcap_\alpha [f_0P_\alpha]_{\omega\alpha Y}=(B),\quad \text{i.e. } (B')=(B). \]

Thus it has been proved that the mappings \(g\) and \(g_0\) coincide. On the basis of Lemma 1,
\[ \bar f_0(\omega X\setminus X)=\omega Y\setminus Y; \]
hence it follows that
\[ g_0(\omega X\setminus X)=\omega\alpha Y\setminus Y. \]
Since \(f=g_0\varphi^{-1}\) and the mapping \(g\) coincides with the mapping \(g_0\), we have
\[ f(\omega\alpha X\setminus X)=\omega\alpha Y\setminus Y. \]
The lemma is completely proved.

For the case of completely regular spaces \(X\) and \(Y\) and their Čech extensions \(\beta X,\beta Y\), this lemma was proved in \((^5)\) (Lemma 1.5).

Definition. The \(\omega\alpha\)-extension \(\omega\alpha X\) of a \(T_1\)-space \(X\) is called perfect if the natural mapping of the Wallman extension \(\omega X\) onto \(\omega\alpha X\) is monotone.

Theorem. Let the extension \(\omega\alpha X\) be perfect, and let the extension \(\omega\alpha Y\) have a punctiform\(^*\) growth. Every perfect mapping
\[ f:X\longrightarrow Y \]
extends to a perfect mapping
\[ \tilde f:\omega\alpha X \longrightarrow \omega\alpha Y. \]

Proof. By Lemma 1, the mapping \(f\) extends to a perfect mapping \(\bar f\) of the space \(\omega X\) onto the space \(\omega Y\), carrying the growth \(\omega X\setminus X\) onto the growth \(\omega Y\setminus Y\). Since \(\omega\alpha X\) is perfect by the hypothesis of the theorem, the natural mapping
\[ \varphi:\omega X \longrightarrow \omega\alpha X \]
is monotone. By definition 2 \((^4)\), there exists a natural mapping
\[ \psi:\omega Y \longrightarrow \omega\alpha Y. \]
Let the point \((A)\in\omega\alpha X\setminus X\); by the monotonicity of \(\varphi\), the set
\[ \varphi^{-1}(A)=A\subset \omega X\setminus X \]
is connected and bicompact. Since \(f\) is perfect, \(\bar f A\) is a connected bicompact subset of the growth \(\omega Y\setminus Y\) (Lemma 3). Hence, from the punctiformity of the growth \(\omega\alpha Y\setminus Y\) it follows that
\[ \psi\bar f A=(B)\in\omega\alpha Y\setminus Y, \]
and thereby the single-valuedness of the mapping
\[ \tilde f=\psi\bar f \]
is proved.

We shall show that
\[ \tilde f\,\omega\alpha X=\omega\alpha Y. \]
Suppose the contrary; then no point is mapped to some point \((B_0)\in\omega\alpha Y\setminus Y\). By the punctiformity of the growth \(\omega\alpha Y\setminus Y\), either the set \(\bar f A\) is entirely contained in some element of the decomposition \(B\subset\omega Y\setminus Y\), or it has no common points with this element of the decomposition. Therefore, for any point \((A)\in\omega\alpha X\setminus X\) we have
\[ \bar f A\cap B_0=\Lambda. \]
In other words, no point of the growth \(\omega X\setminus X\) is mapped onto the set \(B\)—a contradiction to the fact that
\[ \bar f\,\omega X=\omega Y. \]

\(^*\) A space is called punctiform if every connected bicompact subset of it consists of a single point.

The mapping $\tilde f$ is continuous. Indeed, take an arbitrary point $(A_0)\in \omega\alpha X$ and an arbitrary neighborhood $V$ of the point $\tilde f(A_0)$; since the mappings $\psi$ and $\bar f$ are continuous, the set $\bar f^{-1}\psi^{-1}V$ is open in the space $\omega X$. From $\bar f(A_0)\in V$ it follows that
$A_0\subseteq \bar f^{-1}\psi^{-1}(\bar f(A_0))\subseteq \bar f^{-1}\psi^{-1}V$; consequently,
$A_0\subset \bar f^{-1}\psi^{-1}V$.

By Theorem 2 \((^{4})\), $\omega\alpha X$ is a continuous external partition of the space $\omega X$; consequently, in the neighborhood $\bar f^{-1}\psi^{-1}V$ of the element of the partition $A_0$ there exists a marked neighborhood $GA_0$—the set corresponding to it in the space $\omega\alpha X$ will be a neighborhood of the point $(A_0)$, which we denote by $U(A_0)$. By the construction of the neighborhood $U(A_0)$ we have

\[ \tilde f U(A_0)=\psi\bar f\varphi^{-1}U(A_0);\qquad \psi^{-1}U(A_0)=GA_0. \]

By the definition of the neighborhood $U(A_0)$ ($GA_0$ is a marked neighborhood), $GA_0\subseteq \bar f^{-1}\psi^{-1}V$, hence
$\bar f GA_0\subseteq \psi^{-1}V$, $\psi\bar f GA_0\subseteq V$, i.e.
$\psi\bar f\varphi^{-1}U(A_0)\subseteq V$.
The continuity of the mapping $\tilde f$ is proved. Obviously, $\tilde f$ is an extension of the mapping $f$. On the basis of Theorem 7 \((^{4})\), $\tilde f$ is closed; consequently, it is perfect. The theorem is completely proved.

Remark 3. Any homeomorphism $h:\ X \overset{\text{onto}}{\longrightarrow} Y$ ($X,Y$ are $T_1$-spaces), by virtue of our theorem, extends to a perfect mapping $\bar h$ onto the perfect $\omega\alpha$-extensions of the spaces $X$ and $Y$ with punctiform remainder. It is not difficult to see that the mapping $\bar h$ is a homeomorphism of $\omega\alpha X$ onto $\omega\alpha Y$. Consequently, all perfect $\omega\alpha$-extensions with punctiform remainder (if they exist) of a $T_1$-space $X$ coincide with one another up to homeomorphism.

Tiraspol State
Pedagogical Institute

Received
5 III 1965

REFERENCES

1. E. G. Sklyarenko, DAN, 146, No. 5, 1031 (1962).
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4. P. K. Osmatesku, Vestn. Moscow Univ., Mathematics, Mechanics, No. 6, 45 (1963).
5. M. Henriksen, J. Isbell, Duke Math. J., 25, No. 1, 83 (1958).