Full Text
UDC 513.83
MATHEMATICS
N. KROLEVETS
ON LOCALLY PERFECT MAPPINGS
(Presented by Academician P. S. Aleksandrov on 10 VIII 1966)
This paper considers locally perfect mappings, which stand in the same relation to perfect mappings as locally bicompact spaces do to bicompacta. All spaces are assumed to be \(T_1\)-spaces, and all mappings continuous.
A mapping \(f: X \to Y\) will be called locally perfect if for each point \(x\) there exists a neighborhood \(U\) such that the image \(f[U]\) of the closure \(U\) is closed in \(Y\) and \(f|[U]\) is perfect.
Every mapping of a locally bicompact space is locally perfect. In particular, every mapping of a discrete space is locally perfect. Also, if \(f: X \to Y\) is perfect and \(X_0\) is an open set in \(X\), then the restriction \(f|X_0\) is locally perfect.
Let \(f: X \to Y\) be locally perfect. Denote by \(Y^*\) the set of points \(y \in Y\) at which \(f\) is not perfect (i.e., either \(f\) is not closed, or the full preimage \(f^{-1}y\) is not bicompact).
Theorem 1. Let \(f: X \to Y\) be a locally perfect mapping. Then there exist a space \(\widetilde X \supset X\) and a perfect mapping \(\widetilde f: \widetilde X \to Y\) such that: 1) \(\widetilde f|X = f\); 2) \(\widetilde f(\widetilde X \setminus X) = Y^*\); 3) the restriction \(\widetilde f|(\widetilde X \setminus X)\) is one-to-one; 4) \(X\) is dense in \(\widetilde X\); 5) \(X\) is open in \(\widetilde X\). These conditions determine \(\widetilde X\) and \(\widetilde f\) uniquely.
The space \(\widetilde X\) is obtained from \(X\) as follows: to each full preimage of a point \(y \in Y^*\) we add one new point, and to the full preimages of points \(y \in Y \setminus Y^*\) we add nothing. We leave the topology on \(X\) unchanged. If \(x \in \widetilde X \setminus X\), then its neighborhoods will be sets of the form \(U = \widetilde f^{-1}V \setminus F\), where \(V = V_y\) is an arbitrary neighborhood of the point \(y = \widetilde f x\) in \(Y\), \(F \subset X\) and is closed in \(X\), \(fF\) is closed in \(Y\), and the mapping \(f|F\) is perfect.
The mapping \(\widetilde f\) is closed at the points \(y \in Y^*\). Indeed, let \(Q\) be open in \(\widetilde X\), \(\widetilde f^{-1}y \subset Q\). There exists a neighborhood \(U = \widetilde f^{-1}V \setminus F\) of the point \(x_0 = \widetilde f^{-1}y \cap (\widetilde X \setminus X)\), contained in \(Q\). The set \(F_1 = F \setminus (Q \cap X)\) is closed in \(F\), hence \(fF_1\) is closed in \(Y\), and \(y \notin fF_1\). Therefore \(V_1 = V \setminus fF_1\) is a neighborhood of the point \(y\). It is not hard to see that \(\widetilde f^{-1}V_1 \subset Q\). The closedness of \(\widetilde f\) on \(Y \setminus Y^*\) follows from the openness of this set in \(Y\). It is easy to verify that the full preimages \(\widetilde f^{-1}y\), where \(y \in Y^*\), are bicompact. Thus the mapping \(\widetilde f\) is perfect.
In what follows, for a locally perfect mapping \(f: X \to Y\), the symbols \(\widetilde X\) and \(\widetilde f\) will always denote the space and mapping discussed in the theorem.
Let, for example, \(X = \{(x,y): 0 \le x \le 1,\ 0 \le y < 1\}\), \(Y\) be the segment \([0,1]\) of the abscissa axis, and let \(f\) be the vertical projection of \(X\) onto \(Y\). The mapping \(f\) is locally perfect, although it is not closed at any point of the space \(Y\). In this case
\[
\widetilde X = \{(x,y): 0 \le x \le 1,\ 0 \le y \le 1\}
\]
and \(\widetilde f\) is the vertical projection onto \(Y\).
Theorem 2. Let \(f: X \to Y\) be a locally perfect mapping. If \(Y\) is locally bicompact at the point \(y_0\), then \(X\) is locally bicompact at each point \(x \in f^{-1}y_0\). Consequently, if \(Y\) is a locally bicompact space, then \(X\) is also locally bicompact.
However, under passage to the image under a locally perfect mapping, local bicompactness is not preserved.
Theorem 3. Let the mapping \(f:X\to Y\) be locally perfect.
1) If \(X\) and \(Y\) are Hausdorff, then \(\widetilde X\) is Hausdorff; 2) if \(X\) and \(Y\) are completely regular, then \(\widetilde X\) is completely regular; 3) if \(X\) and \(Y\) have countable bases, then \(\widetilde X\) has a countable base.
1) This is proved directly.
2) Let \(B\) be closed in \(\widetilde X\) and \(x_0\notin B\), \(\tilde f x_0\notin \tilde f B\). Then the existence of a function continuous on \(\widetilde X\), equal to zero at the point \(x_0\) and to one on \(B\), follows from the complete regularity of \(Y\). If \(y_0=\tilde f x_0\in \tilde f B\), \(x_0\in \widetilde X\setminus X\), then, since \(X\) is completely regular and the set \(A=B\cap \tilde f^{-1}y_0\) is bicompact, there exists a neighborhood \(U\), \([U]_{\widetilde X}\subset X\), of the set \(A\) and a function \(\psi\), continuous on \(X\), such that \(\psi|_{X\setminus U}\equiv 0\) and \(\psi\equiv 1\) in some smaller neighborhood \(W\), \([W]_X\subset U\), of the set \(A\). Then the function \(\varphi_1\), \(\varphi_1(x)=\psi(x)\) if \(x\in U\), and \(\varphi_1(x)=0\) if \(x\in \widetilde X\setminus U\), is continuous on \(\widetilde X\), \(\varphi_1(x_0)=0\), and \(\varphi_1|_W\equiv 1\). Since \(B_1=B\setminus W\) is closed in \(\widetilde X\), and \(y_0\notin \tilde f B_1\), by what has been proved there exists a continuous function \(\varphi_2\), \(\varphi_2(x_0)=0\), \(\varphi_2|_{B_1}\equiv 1\). Then the function \(\varphi=\max(\varphi_1,\varphi_2)\) is continuous on \(\widetilde X\) and separates the point \(x_0\) and the set \(B\). Finally, if \(x_0\in X\) and \(\tilde f x_0\in \tilde f B\), then the assertion follows at once from the complete regularity of \(X\).
3) Let \(\{U_n\}\) be a countable base of \(X\) and \(\{V_m\}\) a countable base of \(Y\). Then a countable base of \(\widetilde X\) is formed by \(\{U_n\}\) and all sets of the form
\[ U'_{m,k}=\tilde f^{-1}V_m\setminus \bigcup_{i=1}^{k}[U_i]. \]
Corollary. Let \(f:X\to Y\) be a locally perfect mapping of a Hausdorff space \(X\) onto a regular space \(Y\). Then \(X\) is regular.
However, there exist examples of a locally perfect mapping of a completely regular, but not normal, space onto a normal space.
Theorem 4. If \(X\) and \(Y\) are metric spaces with countable bases and the mapping \(f:X\to Y\) is locally perfect, then the space \(\widetilde X\) is metrizable.
Without the assumption of a countable base, the theorem is false.
Example. The space \(X\) is the interval \([0,1]\) with the discrete topology, \(Y\) is the same interval with the usual topology, and \(f\) is the identity mapping of \(X\) onto \(Y\). Then \(f\) is locally perfect, the space \(X\) is metric, and \(Y\) is even compact. The extended space \(\widetilde X\) is bicompact, but does not have a countable base, and therefore \(\widetilde X\) is not metrizable.
Under locally perfect mappings, metrizability, generally speaking, is preserved neither under passage to the image nor under passage to the preimage.
Theorem 5. Let \(X\) and \(Y\) be metric spaces with countable bases, with \(Y\) an absolute \(G_\delta\), and let the mapping \(f:X\to Y\) be locally perfect. Then \(X\) is also an absolute \(G_\delta\).
Indeed, the extended space \(\widetilde X\) is an absolute \(G_\delta\) ((1), Theorem 4), and since \(X\) is open in \(\widetilde X\), \(X\) is also an absolute \(G_\delta\).
Theorem 6. Let \(X\) and \(Y\) be finite-dimensional metric spaces with countable bases and let the mapping \(f:X\to Y\) be locally perfect. If \(\dim X-\dim Y=k\), then there exists a point of the space \(Y\) whose full preimage has dimension \(\ge k\).
Let \(\tilde f:\widetilde X\to Y\) be a perfect extension of \(f\). Since \(\dim X\le \dim \widetilde X\), there exists a point \(y_0\in Y\) such that \(\dim \tilde f^{-1}y_0\ge k\). Then also \(\dim f^{-1}y_0\ge k\).
In what follows, the spaces under consideration will be assumed Hausdorff.
Let \(f\) be a mapping of \(X\) onto \(Y\). If \(X_1 \supset X\), \(X\) is dense in \(X_1\), and \(f_1\) is a perfect mapping of \(X_1\) onto \(Y\) which coincides with \(f\) on \(X\), then \(f_1\) will be called a perfect extension of the mapping \(f\).
Theorem 7. Let \(f:X\to Y\) be a locally perfect mapping and let \(\tilde f_1:\tilde X_1\to Y\) be any perfect extension of it, with \(\tilde X_1\) Hausdorff. Then: 1) \(X\) is open in \(\tilde X_1\); 2) if \(y\in Y\setminus Y^*\), then \(\tilde f_1^{-1}y=f^{-1}y\), where \(Y^*\) is the set of points \(y\in Y\) at which \(f\) is not perfect.
The proof is based on the following lemma.
Lemma. Let \(\varphi:X\to Y\) be a perfect mapping. If \(A\subset X\), \(\varphi|A\) is perfect, and \(\varphi A\) is closed in \(Y\), then \(A\) is closed in \(X\).
This follows easily from (2), Lemma 1.4.
Proof of Theorem 7. 1) Since \(f\) is locally perfect, each point \(x\) has in \(X\) a neighborhood \(U\) such that \([U]_X\) is closed in \(\tilde X_1\) (by the lemma). There exists a neighborhood \(V_x\) in \(\tilde X_1\) such that \(U=X\cap V\). Then \((V\setminus [U]_X)\cap X=\Lambda\), and since \(X\) is dense in \(\tilde X_1\), \(V\setminus [U]_X\) is empty. Hence \(V\subset [U]_X\), i.e. \(V\subset X\), \(U=V\).
2) Let \(x_0\in \tilde f_1^{-1}y\setminus f^{-1}y\), \(y\in Y\setminus Y^*\). Since \(y\in Y\setminus Y^*\), there exists a neighborhood \(U[f^{-1}y]\) such that \([U]_X\) is closed in \(\tilde X_1\), and a neighborhood \(V_y\) such that \(f^{-1}V\Subset U\). \((\tilde f_1^{-1}V\setminus f^{-1}V)\cap X=\Lambda\), then \(W=\tilde f_1^{-1}V\setminus [U]_X\) is nonempty, open in \(\tilde X_1\), and \(W\cap X=\Lambda\), which is impossible.
If the mapping \(f:X\to Y\) is locally perfect, then the perfect extension \(\tilde f\) and the space \(\tilde X\) constructed in Theorem 1 are minimal in the following natural sense.
Theorem 8. Let \(\tilde f_1:\tilde X_1\to Y\) be any perfect extension of the locally perfect mapping \(f:X\to Y\). Then there exists a continuous mapping \(\tilde h:\tilde X_1\to \tilde X\) such that \(\tilde h|x|=x\), if \(x\in X\), and \(\tilde f_1=\tilde f\tilde h\).
Define \(\tilde h\) as follows. If \(x\in X\), then \(\tilde h|x|=x\). If, however, \(x\in\tilde X_1\setminus X\), then, by Theorem 7, there exists a unique point \(x'\in\tilde X\setminus X\) such that \(\tilde f_1x=\tilde f x'\), and then we set \(\tilde h|x|=x'\in\tilde X\setminus X\). It is easy to see that \(\tilde h\) is continuous on \(\tilde X_1\).
Now we restrict ourselves to considering completely regular spaces. Every continuous mapping \(f:X\to Y\) extends uniquely to a continuous mapping \(f_B:\beta X\to BY\), where \(BY\) is any bicompact extension of the space \(Y\). The mapping \(f_B\) is perfect on the set \(f_B^{-1}Y\). Denote \(f_B^{-1}Y\) by \(\tilde X_0\); \(\tilde X_0\) is completely regular. Let \(f_B|\tilde X_0=\tilde f_0\). The space \(X\) is contained in \(\tilde X_0\) and is dense in it; hence the mapping \(\tilde f_0:\tilde X_0\to Y\) is a perfect extension of the mapping \(f:X\to Y\). The extension \(\tilde f_0\) and the space \(\tilde X_0\) are maximal, namely:
Theorem 9. Let \(\tilde f_1:\tilde X_1\to Y\) be any perfect extension of the mapping \(f:X\to Y\), where \(X\), \(Y\), and \(\tilde X_1\) are completely regular. Then there exists a continuous mapping \(\tilde h:\tilde X_0\to\tilde X_1\) such that \(\tilde h|x|=x\), if \(x\in X\), and \(\tilde f_0=\tilde f_1\tilde h\).
We shall call a mapping \(f:X\to Y\) locally closed if every point \(x\in X\) has a neighborhood \(U\) such that \(f[U]\) is closed in \(Y\), the mapping \(f|[U]\) is closed, and \([U]\cap \mathrm{Fr}\, f^{-1}y\) is a bicompactum for any point \(y\in f[U]\).
Theorem 10. Let \(f\) be a locally closed mapping of a metric space \(X\) onto a space \(Y\) satisfying the first axiom of countability. Then there exist a space \(\tilde X\supset X\) and a closed mapping \(\tilde f:\tilde X\to Y\) such that: 1) \(\tilde f|X=f\); 2) \(\tilde f(\tilde X\setminus X)=Y^*\), where \(Y^*\) is the set of points \(y\in Y\) at which \(f\) is not closed; 3) \(\tilde f|(\tilde X\setminus X)\) is one-to-one; 4) \(X\) is dense in \(\tilde X\); 5) \(X\) is open in \(\tilde X\). The conditions 1, 2, and 3 determine \(\tilde X\) and \(\tilde f\) uniquely.
The space \(\tilde X\) is obtained from \(X\) as follows: to each full inverse image of a point \(y\in Y^*\) we add one new point, and to the full inverse images of points \(y\in Y\setminus Y^*\) we add nothing. We leave the topology on \(X\) unchanged. If \(x\in\tilde X\setminus X\), its neighborhoods will be the sets
of the form \(U=\tilde f^{-1}V\setminus F\), where \(V=V_y\) is an arbitrary neighborhood of the point \(y=\tilde f x\) in \(Y\), \(F\subset X\) and is closed in \(X\), \(fF\) is closed in \(Y\), the mapping \(f|F\) is closed, and \(F\cap \operatorname{Fr} f^{-1}y\) is bicompact for \(y\in fF\).
Analogously to Theorem 1, the closedness of \(\tilde f\) is proved.
If the space \(Y\) is Hausdorff (respectively regular), then \(\tilde X\) is also Hausdorff (respectively regular).
Theorem 11. Let \(f\) be a locally closed mapping of a normal space \(X\) onto a space \(Y\) with the first axiom of countability. Then the boundary of the full preimage \(f^{-1}y\) of each point \(y\) is locally compact.
If \(f\) is a locally perfect mapping of a metrizable space \(X\) onto a space \(Y\) with the first axiom of countability, then, extending it by the method described in Theorem 10, we obtain a closed but, generally speaking, not perfect extension.
I take this opportunity to express my deep gratitude to my adviser I. A. Vainshtein for constant assistance and valuable advice, which had a great influence on the direction of this work.
Moscow State University
named after M. V. Lomonosov
Received
8 VII 1966
REFERENCES
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- M. Henriksen, J. Isbell, Duke Math. J., 25, No. 1, 83 (1958).