Let \(p: X \to Y\) be a closed continuous surjective map such that \(p^{-1}(\ts{y})\) is compact for each \(y \in Y\). (Such a map is called a perfect map.)
Show that if \(X\) is Hausdorff, then so is \(Y\).
Show that if \(X\) is regular, then so is \(Y\).
Show that if \(X\) is locally compact, then so is \(Y\).
Show that if \(X\) is second-countable, then so is \(Y\). [Hint: Let \(\mathcal{B}\) be a countable basis for \(X\). For each finite subset \(J\) of \(\mathcal{B}\), let \(U_J\) be the union of all sets of the form \(p^{-1}(W)\), for \(W\) open in \(Y\), that are contained in the union of the elements of \(J\).]
Solution
We repeatedly use the following consequence of closedness of \(p\): if \(U\subset X\) is open and contains the fiber \(p^{-1}(y)\), then \[ W=Y\setminus p(X\setminus U) \] is an open neighborhood of \(y\) and \(p^{-1}(W)\subset U\).
Let \(y_1\ne y_2\). The compact fibers \(F_i=p^{-1}(y_i)\) are disjoint compact subsets of Hausdorff \(X\), so they have disjoint open neighborhoods \(U_1,U_2\). Applying the lemma gives neighborhoods \(W_i\ni y_i\) with \(p^{-1}(W_i)\subset U_i\). Then \(W_1\cap W_2=\varnothing\). Hence \(Y\) is Hausdorff.
Let \(C\subset Y\) be closed and \(y\notin C\). The compact fiber \(F=p^{-1}(y)\) is disjoint from the closed set \(A=p^{-1}(C)\). Since \(X\) is regular, each point of \(F\) has an open neighborhood disjoint from an open neighborhood of \(A\); compactness of \(F\) and a finite intersection argument give disjoint open sets \(U,V\subset X\) with \(F\subset U\) and \(A\subset V\). The lemma gives an open \(W\ni y\) with \(p^{-1}(W)\subset U\). For each \(c\in C\), apply the lemma to the fiber \(p^{-1}(c)\subset V\), and take the union of the resulting neighborhoods to obtain an open \(O\supset C\) with \(p^{-1}(O)\subset V\). Then \(W\cap O=\varnothing\). Thus \(Y\) is regular.
Let \(y\in Y\). Compactness of \(F=p^{-1}(y)\) and local compactness of \(X\) give finitely many neighborhoods \(U_1,\dots,U_n\) of points of \(F\), each contained in a compact subspace \(K_i\subset X\), with \(F\subset U=\bigcup_iU_i\). Then \(K=K_1\cup\cdots\cup K_n\) is compact and contains \(U\). By the lemma there is a neighborhood \(W\ni y\) with \(p^{-1}(W)\subset U\subset K\). Hence \[ W\subset p(K), \] and \(p(K)\) is compact. Thus \(Y\) is locally compact.
Let \(\mathcal B\) be a countable basis for \(X\). For each finite \(J\subset\mathcal B\), let \[ U_J=\bigcup\{p^{-1}(W): W\subset Y\text{ open and }p^{-1}(W)\subset\bigcup J\}. \] Then \(U_J\) is open and saturated, so \(p(U_J)\) is open in \(Y\). There are only countably many finite \(J\), hence only countably many sets \(p(U_J)\).
We show they form a basis. Let \(y\in W\) with \(W\subset Y\) open. The compact fiber \(F=p^{-1}(y)\) is contained in the open set \(p^{-1}(W)\). Choose finitely many basis elements \(B_1,\dots,B_n\in\mathcal B\) such that \[ F\subset B_1\cup\cdots\cup B_n\subset p^{-1}(W). \] Let \(J=\{B_1,\dots,B_n\}\). By the closed-map lemma there is an open neighborhood \(W'\ni y\) with \[ p^{-1}(W')\subset\bigcup J. \] Hence \(y\in p(U_J)\). Also by definition \(U_J\subset p^{-1}(W)\), since every saturated open subset counted in \(U_J\) lies in \(\bigcup J\subset p^{-1}(W)\). Thus \[ y\in p(U_J)\subset W. \] Therefore \(Y\) is second countable.