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54 A first course in algebraic topology If x1 = x2 then Yi * Y2 and a similar argument shows that there are disjoint opensetsXX V1 andXXV2inXX Ywithw1EXX Vtandw2EXXV2. Conversely, if X X Y is Hausdorff then so are the subspaces X X f y }, {x} XYandhencesoareXandY. Thus spaces like S' X S' X ... X S' are Hausdorff. Although subspaces of Hausdorff spaces are Hausdorff and products of Iiausdorff spaces are Hausdorff it is not true in general that a quotient space of a Hausdorff space is Hausdorff. As an example let X be a Hausdorff space with a subset A of X which is not closed, (e.g. X = R, A (O,l)). Let Y be where is the equivalence relation on X given by x x' if and only if c A (intuitively Y is X with A shrunk to a point; in general we denote Y = X/'— by X/A). If we give Y the quotient topology with g: X Y being the natural projection then the inverse image of the point E A is A which is not closed in X. Therefore the J E Y where point [xo] is not closed in Y and Y is not Hausdorff. To ensure that a quotient space Y of a Hausdorff space X is Hausdorff we x = x' or { x,x'} need to impose further restrictions on X. As an example we give the following result. Theorem Let Y be the quotient space of the topological space X determined by the surjective mapping f: X Y. If X is compact Hausdorff and f is closed then Y is (compact) Hausdorff. 8.11 Proof Points of Y are images of points in X which are closed in X, thus the points of Y are closed. Let Yi and Y2 be a pair of distinct points in Y. The are disjoint closed subsets of X. For each point sets (yj) and there are a pair of disjoint open sets xE f'(y1) and each point a E is closed It Is also with xE Ux,a and a E Vx,a. Since and compact and so there is a finite subcover of { Vx,a; a E (Y2) } which In covers f'(y3), say { Vx,a; a E A ) where A is a finite subset of and with xE U, and f1 (y2) ç particular there are disjoint open sets in fact U aEA aEA x E f'(y1) } an open cover of f1(y1) which is compact, x E B } where B is a finite subset of hence there is a finite subcover { Now f Thus the sets is 55 Hausdorff spaces U= U V= C) x€B xEB are disjoint open sets with f is closed, by assumption, f(X - U) and f(X - V) are closed subsets of Y so that W1 = Y - f(X - U) and W3 = Y - f(X - V) are open subsets of Y with Yi E W1 and Y2 E W2. Finally, we just need to check that W1 C) W2 W2, then y f(X - U) and y 0. Suppose therefore that y E W1 C) (X - V) = 0 from f(X - V). Therefore (X - U) = 0 and which it follows that U ('IV =0 and henceW1 flW2 As 8.12 a corollary we get the following result. Corollary If X is a compact Hausdorff G-space with G fInite then X/G is a compact Hausdorff space. Proof 1.et C be a closed subset of X. Then U gC iT gEG where ir: X -+ X/G is the natural projection. Since the action of g E G on X is closed and is a homeomorphism gC is closed for all g G. Thus ii hence ii'(C) is closed which shows that ir is a closed mapping. So, for example, R pa is a compact Hausdorff space. For another corollary to Theorem 8.11 consider a space X with a subset A c X. Recall that X/A denotes Xh where is the equivalence relation on X given by x x' if and only if x = x' or x,x' E A 8.13 Corollary If X is a compact Hausdorff space and A is a closed subset of X then X/A is a compact Hausdorff space. Proof Let C be a closed subset of X and let p: X X/A denote the natural map. IfC C) A = 0 then p(C)Cisclosed. lfCC)A*Øthenp(C)=p(C-A) Up(C ('I A) which Is closed because C U A. Thus pis a closed map. (p(C-A)Up(COA))(C-A)UA = Other restrictions on a Hausdorff space ensuring that a quotient is also Hausdorff appear In the next set of exercises, where a converse to Theorem 8.11 is also given.