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Friday, September 12, 2014

Local Compactness and Products of Quotient Maps (3.29.11)

James Munkres Topology, chapter 3.29, exercise 11:

MathJax TeX Test Page (a) Let Z be a locally compact Hausdorff space, and let p:XY be a quotient map. Show that the map π=p×1:X×ZY×Z is a quotient map.

(b) Let p:AB and q:CD 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 AY×Z is open when π1(A)X×Z is open. For any given y×zA, when x×z is such that π(x×z)=y×z (i.e. xp1(y)) we find a neighborhood U of x saturated with respect to p and a neighborhood V of z such that U×Vπ1(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 π1(A) containing x×z since π1(A) is open. Since Z is locally compact Hausdorff, we may assume V is such that ¯V is compact and UׯVπ1(A).

Given open Ui such that x×zUiׯVπ1(A), we construct Ui+1 such that x×zUi+1ׯVπ1(A) and p1(p(Ui))Ui+1, and we shall set U=Ui. To see this construction, note that for each up1(p(Ui)) we see uׯV is compact, so that by the tube lemma we may choose a neighborhood Vu of u such that VuׯVπ1(A). Letting Ui+1=Vu, we see each of the two desired properties are fulfilled.

We see p1(p(U))=U by the second desired property of each Ui, so that U is saturated. Since also U×VUׯVπ1(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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