UDC 513.831

MATHEMATICS

N. V. VELICHKO

ON THE EXTENSION OF MAPPINGS OF TOPOLOGICAL SPACES

(Presented by Academician P. S. Aleksandrov, February 27, 1967)

The paper is devoted to questions concerning the extension of continuous and $\theta$-continuous (see (1)) mappings of topological Hausdorff spaces to their Hausdorff (mainly $H$-closed) extensions. A space $X$ is called completely separated if any two distinct points of it have disjoint closed neighborhoods. A set $P$ in a space $X$ is called $\delta$-closed if, for every point $x \notin P$, there is a neighborhood $Ox$ such that $\langle [Ox]\rangle \cap P = \varnothing$ (everywhere here $[A]$ denotes the closure, and $\langle A\rangle$ the open kernel of the set $A$) (see (12)). An extension $\rho X$ of a space $X$ is called paracombinatorial if, for every pair of open disjoint sets in $X$, the intersection of their closures in $\rho X$ lies in $X$.

Lemma 1. A Hausdorff extension $\alpha X$ of a Hausdorff space $X$ is $H$-closed if and only if, for every centered system of sets $\{H_\alpha\}$ open in $X$, we have
\[ \bigcap_\alpha [H_\alpha]_{\alpha X} \ne \varnothing. \]

Theorem 1. Every perfect continuous mapping $f$ of a Hausdorff space $X$ onto a Hausdorff space $Y$ extends, with preservation of continuity, to a mapping $\hat f$ of the Katětov extension $\tau X$ onto an arbitrary $H$-closed extension $cY$ of the space $Y$ in such a way that
\[ \hat f(\tau X \setminus X)=cY \setminus Y. \]

Proof. Let $\xi \in \tau X \setminus X$, let $\{H'_\alpha\}_{\alpha\in I}$ be the system of all its open neighborhoods in $\tau X$, and let $\{H_\alpha\}=\{H'_\alpha\cap X\}$. Then
\[ \bigcap_\alpha [H_\alpha]_X=\varnothing. \]
In each set $H_\alpha$ choose a point $x_\alpha$. Let $B_\alpha=f^{-1}fx_\alpha$. The set $B_\alpha$ is bicompact. Denote by $\{O_{\gamma_\alpha}B_\alpha\}_{\gamma_\alpha\in\Gamma_\alpha}$ the system of all open neighborhoods in $X$ of the bicompactum $B_\alpha$. Construct the systems $\{O_{\gamma_\alpha}B_\alpha\}$ for each $\alpha\in I$. Consider the system
\[ \{T_{\gamma_\alpha,\alpha}\}=\{H_\alpha\cup O_{\gamma_\alpha}B_\alpha\}, \]
where $\gamma_\alpha$ runs through $\Gamma_\alpha$, and $\alpha$ runs through $I$. We shall prove that
\[ \bigcap_{\gamma_\alpha\in\Gamma_\alpha,\ \alpha\in I} [T_{\gamma_\alpha,\alpha}]_{\tau X}=\{\xi\}. \]
Let a point $\eta\ne \xi$ be arbitrary, and let $O\eta$ and $O\xi$ be disjoint (open) neighborhoods of the points $\eta$ and $\xi$ in $\tau X$. Let $O\xi\cap X=H_\alpha$. Since $B_\alpha$ is bicompact, the point $\eta$ has a neighborhood $V\eta$ such that $[V\eta]\cap B_\alpha=\varnothing$. Then
\[ V\eta\cap O\eta\cap T_{\gamma_\alpha,\alpha}=\varnothing, \]
where
\[ T_{\gamma_\alpha,\alpha}=H_\alpha\cup \bigl(X\setminus [V\eta\cap X]_X\bigr). \]
Consequently,
\[ \eta\notin \bigcap [T_{\gamma_\alpha,\alpha}]_{\tau X} \quad\text{and}\quad \bigcap [T_{\gamma_\alpha,\alpha}]_{\tau X}=\{\xi\}. \]

The set $X\setminus T_{\gamma_\alpha,\alpha}$ is closed in $X$; therefore the closed set
\[ S_{\gamma_\alpha,\alpha}=f(X\setminus T_{\gamma_\alpha,\alpha}) \]
does not cover $Y$, since $fx_\alpha\notin S_{\gamma_\alpha,\alpha}$. Put
\[ L_{\gamma_\alpha,\alpha}=Y\setminus S_{\gamma_\alpha,\alpha}. \]
The family
\[ t=\{L_{\gamma_\alpha,\alpha}\}_{\gamma_\alpha\in\Gamma_\alpha,\ \alpha\in I} \]
of open sets in $Y$ is centered. Indeed, consider an arbitrary finite collection of sets from $t$:
\[ \{L_{\gamma_{\alpha_i},\alpha_i}\},\quad i=1,2,\ldots,n. \]
Let
\[ H_\beta=\bigcap_{i=1}^n H_{\alpha_i}. \]
Then
\[ fx_\beta\in \bigcap_{i=1}^n L_{\gamma_{\alpha_i},\alpha_i} \quad\text{and}\quad \bigcap_{i=1}^n L_{\gamma_{\alpha_i},\alpha_i}\ne\varnothing. \]
From the centeredness of the system $\{L_{\gamma_\alpha,\alpha}\}$ it follows that
\[ \bigcap_{\gamma_\alpha,\alpha}[L_{\gamma_\alpha,\alpha}]_{cY}=B\ne\varnothing \]
(Lemma 1). We shall prove,

that the set \(B\) reduces to a single point \(y\). Suppose that \(B\) contains two points \(y_1\) and \(y_2\). Let \(Oy_1\) and \(Oy_2\) be their disjoint open neighborhoods. Then
\[ f^{-1}(Oy_1\cap Y)\cap f^{-1}(Oy_2\cap Y)\ne\varnothing, \]
but
\[ T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_1\cap Y)\ne\varnothing \quad\text{and}\quad T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_2\cap Y)\ne\varnothing \]
for arbitrary \(\gamma_\alpha\) and \(\alpha\).

We shall prove that: (A) for any open neighborhood \(Oy\) of an arbitrary point \(y\in B\) we have
\[ \xi\in [f^{-1}(Oy\cap Y)]_{\tau X}. \]
Indeed, suppose the contrary: assume that the point \(\xi\) has an open neighborhood \(O\xi\) not intersecting \(f^{-1}Oy\). Let \(O\xi\cap X=H_\alpha\). Then \(fx_\alpha\ne y\). Let \(Vy\) and \(Vfx_\alpha\) be disjoint open neighborhoods of the points \(y\) and \(fx_\alpha\) in \(cY\). Then, for
\[ T_{\gamma_\alpha,\alpha}=H_\alpha\cup f^{-1}(Vfx_\alpha) \]
and
\[ Wy=Vy\cap Oy, \]
we would have
\[ Wy\cap L_{\gamma_\alpha,\alpha}=\varnothing, \]
which is impossible. Thus (A) is proved. Consequently,
\[ \xi\in [T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_1\cap Y)]_{\tau X} \cap [T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_2\cap Y)]_{\tau X}, \]
where \(T_{\gamma_\alpha,\alpha}\) is chosen arbitrarily. But the sets
\[ T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_1\cap Y)=M_1 \]
and
\[ M_2=T_{\gamma_\alpha,\alpha}\cap f^{-1}(Oy_2\cap Y) \]
are open in \(X\) and do not intersect. Therefore (since \(\tau X\) is a paracombinatorial extension)
\[ [M_1]_{\tau X}\cap [M_2]_{\tau X}\cap(\tau X\setminus X)=\varnothing. \]
We have arrived at a contradiction. Hence \(B=\{y\}\).

Put \(\hat f\xi=y\), and \(\hat f x=fx\) for points \(x\in X\). We obtain a mapping \(\hat f\) of the space \(\tau X\) into the space \(cY\), which extends \(f\), and
\[ \hat f(\tau X\setminus X)\subseteq cY\setminus Y. \]
The latter inclusion follows from the fact that the point \(\hat f\xi=y\in cY\setminus Y\), for for any point \(\bar y\in Y\) the set \(B\bar y=f^{-1}\bar y\) and the point \(\xi\in\tau X\setminus X\) have disjoint open neighborhoods \(OB\bar y\) and \(O\xi\), and then, if \(O\xi\cap X=H_\alpha\), we have
\[ \bar y\notin L_{\gamma_\alpha,\alpha} \]
for some \(\gamma_\alpha\).

It is clear that the mapping \(\hat f\) is continuous at each point \(x\in X\). Let \(\xi\in\tau X\setminus X\), \(y=\hat f\xi\), and let \(Oy\) be an arbitrary open neighborhood of the point \(y\) in \(cY\), \(T=Oy\cap Y\). Then
\[ \xi\in [f^{-1}T]_{\tau X} \]
(see (A)), and since \(f^{-1}T\) is open in \(X\), it follows that
\[ \{\xi\}\cup f^{-1}T=O\xi \]
is a neighborhood of the point \(\xi\) such that
\[ \hat f\,O\xi\subseteq Oy. \]
Consequently, \(\hat f\) is continuous on \(\tau X\), and since
\[ \hat f\tau X\supseteq Y, \]
we have
\[ \hat f\tau X=cY, \]
and
\[ \hat f(\tau X\setminus X)=cY\setminus Y. \]
The theorem is proved.

Corollary. The Katětov extension \(\tau X\) of a Hausdorff space \(X\) is mapped continuously and fixing the points of \(X\) onto every \(H\)-closed extension \(cX\) of the space \(X\), and the remainder \(\tau X\setminus X\) is mapped onto the remainder \(cX\setminus X\).

Theorem 2. If \(f\) is a perfect continuous mapping of a Hausdorff space \(X\) onto an \(H\)-closed space \(Y\), then \(X\) is also \(H\)-closed.

Proof. The mapping \(f:X\to Y\) can be extended, preserving continuity, to a mapping
\[ \hat f:\tau X\to Y \]
(Theorem 1), but a perfect mapping is absolutely closed (see (2)), which entails the equality
\[ \tau X=X. \]
Consequently, the space \(X\) is \(H\)-closed. The theorem is proved.

Definition. A mapping \(f\) of a space \(X\) onto a space \(Y\) is called \(\delta\)-continuous if, for every point \(x\in X\) and every neighborhood \(Ou\) of the point \(y=fx\in Y\), there is a neighborhood \(Ox\) of the point \(x\) such that
\[ fOx\subseteq \langle[Oy]\rangle . \]

Lemma 2. A mapping \(f:X\to Y\) is \(\delta\)-continuous if and only if, for every set \(H\) open in \(Y\), the set
\[ f^{-1}(\langle[H]\rangle) \]
is open in \(X\).

Theorem 3. A \(\theta\)-continuous mapping \(f\) of a Hausdorff space \(X\) onto a Hausdorff space \(Y\) is \(\theta\)-continuous in each of the following cases: (a) the space \(X\) is \(H\)-closed and completely regular; (b) the mapping \(f\) is open.

Proof. (a). The space \(Y\) is \(H\)-closed and completely regular. Let \(x\) be an arbitrary point of \(X\), and let \(Oy\) be an arbitrary neighborhood of the point \(y=fx\). We shall prove that the point \(y\) has a neighborhood \(Vy\) such that
\[ [Vy]\subseteq \langle[Oy]\rangle . \]
The set
\[ S=Y\setminus \langle[Oy]\rangle=[Y\setminus [Oy]] \]
is \(H\)-closed as the closure of an open set. For each point \(z\in S\)

and the point \(y\) have open neighborhoods \(O z\) and \(O z y\) such that \([O z]\cap[O z y]=\varnothing\). There is a finite number of points \(z_i\), \(i=1,2,\ldots,n\), such that \(\bigcup_{i=1}^n[O z_i]\supseteq S\). Put \(V y=\bigcap_{i=1}^n O z_i y\). We have \([V y]\cap S=\varnothing\) (for \([O z_i y]\cap[O z_i]=\varnothing\)), whence \([V y]\subseteq\langle[O y]\rangle\). Let \(O x\) be a neighborhood of the point \(x\) such that \(f O x\subseteq[V y]\). Then \(f O x\subseteq\langle[O y]\rangle\). (a) is proved.

(b). The mapping \(f:X\to Y\) is open. Let \(H\) be an arbitrary open set in \(Y\). Since for each point \(z\in f^{-1}(\langle[H]\rangle)=K\) there is a neighborhood \(O z\) such that \(f O z\subseteq[H]\), the set \(L=\bigcup_{z\in K}O z\) will be an open set such that \(f L=[H]\), \(f L\subseteq\langle[H]\rangle\), \(f L\supseteq\langle[H]\rangle\); consequently, \(f L=\langle[H]\rangle\), \(L=f^{-1}(\langle[H]\rangle)\), and the mapping \(f\) is \(\delta\)-continuous (Lemma 2). (b) is established. The theorem is proved.

Theorem 4. Every open \(\theta\)-continuous mapping \(f\) of a Hausdorff space \(X\) onto a Hausdorff space \(Y\) can be extended, with preservation of \(\theta\)-continuity, to a mapping \(\hat f\) of any paracombinatorial (not necessarily \(H\)-closed) extension \(pX\) into any \(H\)-closed extension \(cY\) of the space \(Y\), in such a way that \(\hat f(pX\setminus X)\subseteq cY\setminus Y\).

Proof. Let \(p\in pX\setminus X\) be arbitrary, and let \(\{H_\alpha'\}_{\alpha\in I}\) be the system of all its open neighborhoods in \(pX\), \(\{H_\alpha\}=\{H_\alpha'\cap X\}\). The family \(\{f H_\alpha\}\) of open sets in \(Y\) is centered; therefore \(B=\bigcap_\alpha [f H_\alpha]_{cY}\ne\varnothing\) (Lemma 1). The set \(B\) reduces to a single point \(\eta\). Suppose the contrary—that \(B\) contains two points \(\eta_1\) and \(\eta_2\). Let \(O\eta_1\) and \(O\eta_2\) be disjoint open neighborhoods of the points \(\eta_1\) and \(\eta_2\). In view of the openness of \(f\), the mapping \(f\) is \(\delta\)-continuous (Theorem 3). Then the sets \(L_1=f^{-1}(\langle[O\eta_1\cap Y]\rangle)\) and \(L_2=f^{-1}(\langle[O\eta_2\cap Y]\rangle)\) are open in \(X\) and do not intersect. But \([L_1]_{pX}\cap[L_2]_{pX}=K\ni p\), which contradicts the paracombinatoriality of the extension \(pX\). Hence \(B=\{\eta\}\).

Put \(\hat f p=\eta\), \(\hat f x=f x\) for points \(x\in X\). We shall prove that the mapping \(\hat f\), which is an extension of \(f\), is \(\theta\)-continuous.

(M). Let the set \(A\) be open in \(X\), and let \(z\in[A]_{pX}\setminus[A]_X\). Then \(\xi=\hat f z\in[f A]_{cY}\). Indeed, if the point \(\xi\) had an open neighborhood \(O\xi\) not intersecting \(f A\), then the open sets \(S=f^{-1}(\langle[O\xi\cap Y]\rangle)\) and \(A\) in \(X\) would not intersect, but \([S]_{pX}\cap[A]_{pX}=L\ni z\in pX\setminus X\), which contradicts the paracombinatoriality of \(pX\).

Let the point \(x\in X\) be arbitrary; \(y=\hat f x=f x\); let \(O y\) be an arbitrary open neighborhood of the point \(y\); let \(O x\) be an open neighborhood of the point \(x\) in \(X\) such that \(f O x\subseteq\langle[O y\cap Y]\rangle\); let \(\widehat{O x}\) be the largest open set in \(pX\) that cuts out \(O x\) in \(X\). If \(z\in\widehat{O x}\setminus O x\), then \(z\in[O x]_{pX}\), whence \(\hat f z\in[f O x]_{cY}\subseteq[O y]_{cY}\) (see (M)); consequently, \(\hat f\,\widehat{O x}\subseteq[O y]_{cY}\), i.e. the mapping \(\hat f\) is \(\theta\)-continuous at the point \(x\in X\).

Now let \(p\in pX\setminus X\), \(\eta=\hat f p\in cY\), and let \(O\eta\) be an arbitrary open neighborhood of the point \(\eta\) in \(cY\). The set \(K=f^{-1}(\langle[O\eta\cap Y]\rangle_Y)\) is open in \(X\), and \(p\in[K]_{pX}\). Note that \([K]_{pX}\cup[X\setminus[K]]_{pX}=pX\), and, by the paracombinatoriality of the extension \(pX\), the point \(p\notin[X\setminus[K]]_{pX}\); therefore the point \(p\) has a neighborhood \(O p\) such that \(O p\subseteq[K]_{pX}\). Let the point \(a\in[K]_{pX}\setminus X\) be arbitrary. Then \(b=\hat f a\in[O\eta]_{cY}\) (see (M)); consequently, \(f O p\subseteq[O\eta]_{cY}\), i.e. the mapping \(\hat f\) is \(\theta\)-continuous at the point \(p\in pX\setminus X\) and, thus, \(\theta\)-continuous on \(pX\). It is not hard to verify that if \(p\in pX\setminus X\), then the point \(\eta=\hat f p\in cY\setminus Y\), i.e. \(\hat f(pX\setminus X)\subseteq cY\setminus Y\). The proof is complete.

Theorem 5. In order that a \(\theta\)-continuous mapping \(f\) of an everywhere dense set \(S\) in a space \(X\) into an \(H\)-closed completely Hausdorff space \(Y\) can be extended to all of \(X\) with preservation of \(\theta\)-

continuity, it is necessary and sufficient that, for any two disjoint canonically closed sets \(B_1\) and \(B_2\) in \(Y\), the preimages \(f^{-1}B_1\) and \(f^{-1}B_2\) have disjoint closures in \(X\).

Proof of necessity. The \(\theta\)-continuous mapping \(f:S\to Y\) extends to all of \(X\). The mapping \(f\) is \(\delta\)-continuous (Theorem 3). Let \(B_1\) and \(B_2\) be canonically closed disjoint sets in \(Y\). \(B_1\) and \(B_2\) are \(\delta\)-closed as closures of open sets (Lemma 2 from \((^3)\)). Suppose, for example, that the point \(\xi \in [f^{-1}B_1]_X\). Then \(f\xi \in [B_1]_\delta=B_1\), by the \(\delta\)-continuity of \(f\); therefore \(\xi \notin [f^{-1}B_2]_X\) and
\[ [f^{-1}B_1]\cap [f^{-1}B_2]_X=\varnothing . \]

Proof of sufficiency. Let a point \(x\in X\) be arbitrary, and let \(\mathfrak F\) be the trace on \(S\) of the filter of neighborhoods \(Bx\) of the point \(x\). \(f(\mathfrak F)\) is a filter base in \(Y\), and since \(Y\) is \(H\)-closed, \(f(\mathfrak F)\) has points of \(\delta\)-adherence (Theorem 2 from \((^3)\)). We prove that the set
\[ L=\bigcap [fV_\alpha]_\delta \]
reduces to a single point \(y\) (where \(V_\alpha=S\cap W_\alpha\), and \(W_\alpha\) ranges over \(Bx\)). Let \(y_1\ne y_2\), \(y_1,y_2\in L\). Choose open neighborhoods \(Oy_1\) and \(Oy_2\) of the points \(y_1\) and \(y_2\) such that \([Oy_1]\cap [Oy_2]=\varnothing\). The sets \([Oy_1]\) and \([Oy_2]\) are canonically closed; hence
\[ [f^{-1}([Oy_1])]_X\cap [f^{-1}([Oy_2])]_X=\varnothing . \]
But \(V_\alpha\cap f^{-1}([Oy_1])\ne\varnothing\) and \(V_\alpha\cap f^{-1}([Oy_2])\ne\varnothing\) for every \(V_\alpha\); consequently the point
\[ x\in [f^{-1}([Oy_1])]_X\cap [f^{-1}([Oy_2])] . \]
The contradiction obtained shows that the set \(L\) consists of a single point \(y\). By Lemma 2 from \((^4)\), \(f(\mathfrak F)\) \(\delta\)-converges to the point \(y\) (i.e., contains the nuclei of the closures of all neighborhoods of the point \(y\)).

Put \(\hat f x=y\). It is clear that \(\hat f x=fx\) on the set \(S\). We prove that \(\hat f\) is a \(\theta\)-continuous extension of the mapping \(f\) to \(X\). Consider an arbitrary open neighborhood \(Oy\) of the point \(y=\hat f x\). There exists an open neighborhood \(W_\alpha\) of the point \(x\) such that \(fV_\alpha\subseteq [Oy]\), where \(V_\alpha=S\cap W_\alpha\). For an arbitrary point \(z\in W_\alpha\) we have
\[ \hat f z\in [fV_\alpha]_\delta=[[Oy]]_\delta=[Oy] \]
(for the image of the trace on \(S\) of the filter of neighborhoods of the point \(z\) \(\delta\)-converges to the point \(\hat f z\)), i.e. \(\hat f W_\alpha\subseteq [Oy]\). The theorem is proved.

Corollary 1 (A. D. Taimanov’s theorem). In order that a continuous mapping \(f\) from an everywhere dense subset \(S\) of a space \(X\) into a bicompact space \(Y\) be extendable to all of \(X\) with preservation of continuity, it is necessary and sufficient that, for any two canonically closed disjoint sets \(B_1\) and \(B_2\) in \(Y\), the preimages \(f^{-1}B_1\) and \(f^{-1}B_2\) have disjoint closures in \(X\).

Corollary 2. Every continuous perfect mapping of an everywhere dense subset \(S\) of a Hausdorff space \(X\) into an \(H\)-closed completely nonseparable space \(Y\) extends to a \(\theta\)-continuous mapping of the space \(X\) into the space \(Y\).

Ural State University
named after A. M. Gorky

Received
1 II 1967

REFERENCES

\(^{1}\) S. Fomin, Ann. Math., 44, No. 5, 471 (1943).
\(^{2}\) A. V. Arhangel’skii, Tr. Mosk. matem. obshch., 13, 5 (1964).
\(^{3}\) N. V. Velichko, Matem. sborn., 70 (112), 1, 98 (1966).
\(^{4}\) N. V. Velichko, Sibirsk. matem. zhurn., 6, 1, 64 (1965).