- Lemma. If \(p: X \to Y\) is a quotient map and if \(Z\) is a locally compact Hausdorff space, then the map
\[ \pi = p \times i_Z: X \times Z \to Y \times Z \]
is a quotient map.
[Hint: If \(\pi^{-1}(A)\) is open and contains \(x \times y\), choose open sets \(U_1\) and \(V\) with \(\overline{V}\) compact, such that \(x \times y \in U_1 \times V\) and \(U_1 \times \overline{V} \subset \pi^{-1}(A)\). Given \(U_i \times \overline{V} \subset \pi^{-1}(A)\), use the tube lemma to choose an open set \(U_{i+1}\) containing \(p^{-1}(p(U_i))\) such that \(U_{i+1} \times \overline{V} \subset \pi^{-1}(A)\). Let \(U = \bigcup U_i\); show that \(U \times V\) is a saturated neighborhood of \(x \times y\) that is contained in \(\pi^{-1}(A)\).]
An entirely different proof of this result will be outlined in the exercises of §46.
- Theorem. Let \(p: A \to B\) and \(q: C \to D\) be quotient maps. If \(B\) and \(C\) are locally compact Hausdorff spaces, then \(p \times q: A \times C \to B \times D\) is a quotient map.
Solution
- Let \(\pi=p\times i_Z\), and suppose \(A\subset Y\times Z\) is such that \(\pi^{-1}(A)\) is open. Fix \((y,z)\in A\) and choose \(x\in p^{-1}(y)\). By local compactness and the preceding exercise, choose an open neighborhood \(V\) of \(z\) with compact closure \(\overline V\). Since \(\pi^{-1}(A)\) is open and contains \((x,z)\), shrink to an open \(U_1\ni x\) such that \[ U_1\times\overline V\subset\pi^{-1}(A). \]
Inductively, assume \(U_i\times\overline V\subset\pi^{-1}(A)\). For each \(x'\in p^{-1}(p(U_i))\), the compact slice \(\{x'\}\times\overline V\) lies in the open set \(\pi^{-1}(A)\). By the tube lemma there is an open neighborhood \(W_{x'}\) of \(x'\) with \[ W_{x'}\times\overline V\subset\pi^{-1}(A). \] Let \(U_{i+1}=\bigcup_{x'}W_{x'}\). Then \(U_{i+1}\) is open, contains \(p^{-1}(p(U_i))\), and still satisfies the product inclusion.
Set \(U=\bigcup_iU_i\). Then \(U\) is open and saturated: if \(u\in U_i\), the whole fiber \(p^{-1}(p(u))\subset U_{i+1}\subset U\). Hence \(p(U)\) is open in \(Y\), because \(p\) is quotient and \(U=p^{-1}(p(U))\). Also \[ p(U)\times V\subset A. \] Thus every point of \(A\) has an open neighborhood contained in \(A\), so \(A\) is open. Therefore \(\pi\) is quotient.
- Factor \[ p\times q=(i_B\times q)\circ(p\times i_C). \] Since \(C\) is locally compact Hausdorff, part (a) shows \(p\times i_C:A\times C\to B\times C\) is quotient. Since \(B\) is locally compact Hausdorff, applying part (a) to \(q:C\to D\) gives \(i_B\times q:B\times C\to B\times D\) quotient. A composition of quotient maps is quotient, so \(p\times q\) is quotient.