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 \(Y\) is compact, then \(X\) is compact. [Hint: If \(U\) is an open set containing \(p^{-1}(\ts{y})\), there is a neighborhood \(W\) of \(y\) such that \(p^{-1}(W)\) is contained in \(U\).]
Solution
We first prove the hinted lemma. Let \(U\subset X\) be open with \(p^{-1}(y)\subset U\). Since \(p^{-1}(y)\) is compact and \(U\) already contains the whole fiber, set \(F=X\setminus U\), which is closed. Because \(p\) is closed, \(p(F)\) is closed in \(Y\). Also \(y\notin p(F)\). Therefore \[ W=Y\setminus p(F) \] is an open neighborhood of \(y\), and \(p^{-1}(W)\subset U\).
Now let \(\mathcal U\) be an open cover of \(X\). For each \(y\in Y\), compactness of the fiber \(p^{-1}(y)\) yields finitely many members \(U_{y,1},\dots,U_{y,m_y}\in\mathcal U\) whose union \(U_y\) contains the fiber. By the lemma, there is an open neighborhood \(W_y\) of \(y\) such that \[ p^{-1}(W_y)\subset U_y. \] The sets \(W_y\) cover \(Y\). Since \(Y\) is compact, choose finitely many \(y_1,\dots,y_r\) such that \(Y=W_{y_1}\cup\cdots\cup W_{y_r}\). Then the finitely many members of \(\mathcal U\) appearing in the corresponding finite unions \(U_{y_j}\) cover \(X\). Hence \(X\) is compact.