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Universally Closed Maps
In general, the fact that continuous functions \(f_1:X_1 \rightarrow Y_1\) and \(f_2: X_2 \rightarrow Y_2\) are closed does not imply that their product \(f_1\times f_2: X_1\times X_2 \rightarrow Y_1\times Y_2\) is closed.
Definition 1 A continuous function \(f:X \rightarrow Y\) is called universally closed if for any topological space \(Z\), the map \(f\times\id_Z: X\times Z \rightarrow Y\times Z\) is closed.
Taking \(Z=\{\ast\}\), one can show that any universally closed map is a closed map, but the converse does not hold. However, the following holds.
Proposition 2 If \(f:X \rightarrow Y\) is a continuous injective map, then the following are all equivalent.
- \(f\) is universally closed.
- \(f\) is closed.
- \(f\) is a homeomorphism between \(X\) and \(f(X)\).
Proof
By the above argument, it is obvious that if the first condition holds then the second condition holds. On the other hand, since \(f\) is injective, considering the canonical decomposition of \(f\), we know that it becomes a homeomorphism between \(X\) and \(f(X)\). (§Open and Closed Mappings, ⁋Proposition 5) Now assuming the third condition holds, for any \(Z\), since \(f\times\id_Z\) is a homeomorphism from \(X\times Z\) onto the closed subset \(f(X)\times Z\) of \(Y\times Z\), we obtain the desired result.
However, in general one must check directly whether a function \(f\) is universally closed. The second result of the following proposition shows that this can be examined on the target \(Y\).
Proposition 3 Let a continuous function \(f:X \rightarrow Y\) be given. Then the following hold.
- If \(f\) is universally closed, then for any subset \(A\) of \(Y\), the restriction \(f\vert_{f^{-1}(A)}: f^{-1}(A) \rightarrow A\) is also universally closed.
- Let \((A_i)_{i\in I}\) be a covering of \(Y\) that is either (1) a locally finite closed covering, or (2) such that \((\interior A_i)_{i\in I}\) is an open covering of \(Y\). If each restriction \(f\vert_{f^{-1}(A_i)}\) is universally closed, then so is \(f\).
Proof
First, to prove the first result, let an arbitrary topological space \(Z\) be given. Then for any subset \(A\) of \(Y\),
\[(f\vert_{f^{-1}(A)})\times \id_Z=(f\times\id_Z)\vert_{f^{-1}(A\times Z)}\]holds. Now from the assumption that \(f\) is universally closed, \(f\times\id_Z\) is closed, and therefore \((f\times\id_Z)\vert_{f^{-1}(A\times Z)}\) is also closed.
Now let us prove the second result. If \((A_i)\) satisfying the given condition is given, then \((A_i\times Z)\) also satisfies the same condition. Now if the \(f\vert_{f^{-1}(A_i)}\) are universally closed, then the following functions
\[(f\times\id_Z)\vert_{f^{-1}(A_i\times Z)}\]are closed, and thus \(f\times\id_Z\) is also so. (§Open and Closed Mappings, ⁋Proposition 3)
Also, the following holds.
Proposition 4 For two continuous functions \(f:X \rightarrow Y\) and \(g:Y \rightarrow Z\), the following hold.
- If both \(f\) and \(g\) are universally closed, then so is \(g\circ f\).
- If \(g\circ f\) is universally closed and \(f\) is surjective, then \(g\) is universally closed.
- If \(g\circ f\) is universally closed and \(g\) is injective, then \(f\) is universally closed.
- If \(g\circ f\) is universally closed and \(Y\) is Hausdorff, then \(f\) is universally closed.
Proof
The first three results are all obvious from the following formula
\[(g\circ f)\times\id_Z=(g\times\id_Z)\circ(f\times\id_Z)\]and the results of §Open and Closed Mappings, ⁋Proposition 2.
For the last result, define two functions \(\Gamma_f: X \rightarrow X\times Y\) and \(\Gamma_g: Y \rightarrow Z\times Y\) respectively by
\[\Gamma_f(x)=(x,f(x)),\qquad \Gamma_g(y)=(g(y), y)\]Then the formula
\[((g\circ f)\times\id_Y)\circ\Gamma_f=\Gamma_g\circ f\]holds, as can be easily verified. Here \(\Gamma_f\) and \(\Gamma_g\) are homeomorphisms from \(X\) and \(Y\) onto the graphs of \(f\) and \(g\), respectively. (§Product Spaces, ⁋Corollary 4) Also, from the assumption that \(Y\) is Hausdorff, we know that \(\Gamma(f)\subseteq X\times Y\) is a closed set. (§Hausdorff Spaces, ⁋Corollary 7) Therefore, from Proposition 2, we know that \(\Gamma_f\) is universally closed. On the other hand, since one can show without difficulty that the product of universally closed maps is universally closed, combining this with Proposition 4, we know that \((g\circ f)\times\id_Y\) is universally closed. Therefore, the right-hand side \(\Gamma_g\circ f\) of the above formula is also universally closed, and since \(\Gamma_g\) is injective, \(f\) is universally closed.
Compactness and Universally Closed Maps
So far, it is not visible how this definition is related to compactness; in this section we examine the relationship between them. Before that, we need to say a little about filters in order to use §Compactness, ⁋Lemma 1.
Let an arbitrary topological space \(X\) and an arbitrary filter \(\mathcal{F}\) on it be given. Consider the set \(X'=X\cup \{\ast_X\}\) obtained by adding one point to \(X\), and the filter
\[\mathcal{F}'=\{F\cup\{\ast_X\}: F\in \mathcal{F}\}\]on it. Now defining \(\mathcal{N}(x)=\uparrow\{x\}\) for any \(x\in X'\) other than \(\ast_X\), and \(\mathcal{N}(\ast_X)=\mathcal{F}'\), this satisfies all four conditions of §Open Sets, ⁋Proposition 6, and thus a topology on \(X'\) is defined. In this topological space \(X'\), \(\ast_X\) is contained in the closure of \(X\), and it is obvious that \(\mathcal{F}=\mathcal{F}'\vert_X=\mathcal{N}(\ast_X)\vert_X\).
Then we can prove the following lemma.
Lemma 5 For a topological space \(X\), suppose \(f: X \rightarrow \{\ast\}\) is universally closed. Then \(X\) is compact.
Proof
For an arbitrary filter \(\mathcal{F}\) on \(X\), consider the space \(X'\) obtained from the above argument. Also, define the subset \(\Delta\) of \(X\times X'\) by
\[\Delta=\{(x,x)\mid x\in X\}\]Then we can consider the closure \(\cl\Delta\) of \(\Delta\), and from the assumption that \(f\) is universally closed, we know that the image of \(\cl\Delta\) under \(f\times\id_{X'}:X\times X'\rightarrow \{\ast\}\times X'\cong X'\) is a closed set. Now since this image contains \(x\), from the assumption that \(\ast_X\) is contained in the closure, we know that there exists a suitable \(x\in X\) such that \((x,\ast_X)\in \cl\Delta\). Then \(x\) is a cluster point of \(\mathcal{F}\), and therefore, considering an ultrafilter containing \(\mathcal{F}\), we know that \(x\) is the limit point of that filter.
The converse of the above lemma also holds. More generally, the following holds.
Theorem 6 For a continuous function \(f:X \rightarrow Y\), the following are equivalent.
- \(f\) is universally closed.
- \(f\) is closed, and for each \(y\in Y\), the fiber \(f^{-1}(y)\) is compact.
Proof
If the first condition holds, then for any \(y\in Y\), we know that \(f\vert_{f^{-1}(y)}\) is universally closed, and from the preceding lemma we know that \(f^{-1}(y)\) is compact. The converse can be proved using the fact that the product of universally closed maps is universally closed.
Therefore, we could have taken the definition of a compact space to be that \(f:X \rightarrow \{\ast\}\) is universally closed. In particular, the following holds.
Corollary 7 If a continuous function \(f:X \rightarrow Y\) is universally closed, then for any compact subset \(C\subseteq Y\), the preimage \(f^{-1}(C)\) is also compact.
Proof
Since \(f\) is universally closed, \(f\vert_{f^{-1}(C)}\) is universally closed. On the other hand, \(C \rightarrow\{\ast\}\) is universally closed from the assumption that \(C\) is compact, and therefore the composition \(f^{-1}(C) \rightarrow C \rightarrow \{\ast\}\) is universally closed, so \(f^{-1}(C)\) is also compact.
Proper Maps
A function \(f\) satisfying the result of Corollary 7 is called a proper map. The following proposition shows that the converse of the above corollary also holds in a special case.
Proposition 8 Let a continuous function \(f:X \rightarrow Y\) between Hausdorff spaces be given, and additionally assume that \(Y\) is locally compact. Then \(f\) being universally closed and \(f\) being proper are equivalent.
Proof
As mentioned earlier, that \(f\) being universally closed implies \(f\) is proper is the result of Corollary 7. Therefore, the core of this proposition is the reverse direction. Since \(Y\) is locally compact, there exists an open covering \((U_i)\) of \(Y\) consisting of open sets each contained in a suitable compact set. Then the \(f^{-1}(\cl U_i)\) are compact in \(X\) and each \(f\vert_{f^{-1}(\cl U_i)}\) is universally closed. Now from Proposition 3, we obtain the desired result.
In particular, applying this to the one-point compactification examined earlier gives the following result.
Corollary 9 Let two locally compact Hausdorff spaces \(X_1, X_2\) be given, and let their one-point compactifications \(\overline{X}_i=X\cup \{\ast_i\}\) be given. Then \(f:X_1: X_2\) being universally closed is equivalent to \(\overline{f}:\overline{X}_1 \rightarrow \overline{X}_2\) defined by the formula
\[\overline{f}(x)=\begin{cases}\ast_2&\text{if $x=\ast_1$}\\f(x)&\text{otherwise}\end{cases}\]being continuous.
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