Download opensetsXX V1 andXXV2inXX Ywithw1EXX Vtandw2EXXV2. {x

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.