(b) Let p:A→B and q:C→D be quotient maps, and let B and C be locally compact Hausdorff. Show p×q is a quotient map.
Proof: (a) It is clear that π is continuous, so it suffices to show A⊆Y×Z is open when π−1(A)⊆X×Z is open. For any given y×z∈A, when x×z is such that π(x×z)=y×z (i.e. x∈p−1(y)) we find a neighborhood U of x saturated with respect to p and a neighborhood V of z such that U×V⊆π−1(A), so that π(U×V)=p(U)×V will be a neighborhood of y×z contained in A. In this fashion, let U1×V be a basis element in π−1(A) containing x×z since π−1(A) is open. Since Z is locally compact Hausdorff, we may assume V is such that ¯V is compact and UׯV⊆π−1(A).
Given open Ui such that x×z∈UiׯV⊆π−1(A), we construct Ui+1 such that x×z∈Ui+1ׯV⊆π−1(A) and p−1(p(Ui))⊆Ui+1, and we shall set U=∪Ui. To see this construction, note that for each u∈p−1(p(Ui)) we see uׯV is compact, so that by the tube lemma we may choose a neighborhood Vu of u such that VuׯV⊆π−1(A). Letting Ui+1=∪Vu, we see each of the two desired properties are fulfilled.
We see p−1(p(U))=U by the second desired property of each Ui, so that U is saturated. Since also U×V⊆UׯV⊆π−1(A) we have demonstrated the proper U and V, so that π is a quotient map.
(b) This is evident by considering the composition of quotient maps p×q=(1×q)∘(p×1). ◻
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