Page 109 - DMTH503_TOPOLOGY

Unit 10: The Quotient Topology

Notes
2
Example 4: Let X be the closed unit ball {x × y | x + y  1} in , and let X* be the partition
2
2
2
2
of X consisting of all the one-point sets {x × y} for which x + y < 1, along with the set
2
S = {x × y} | x + y = 1}. One can show that X* is homeomorphic with the subspace of 3 called
2
1
the unit 2-sphere, defined by
2
2
S = {(x, y, z) | x + y + z = 1}
2
2
Theorem 1: Let p : X  Y be a quotient map; let A be a subspace of X that is saturated with respect
to p; let q : A  p(A) be the map obtained by restricting p.
(1) If A is either open or closed in X, then q is a quotient map.
(2) If p is either an open map or a closed map, then q is a quotient map.
Proof: Step (1): We verify first the following two equations:
–1
–1
q (V) = p (V) if V  p(A);
p(U A) = p(U) p(A) if U  X
To check the first equation, we note that since V  p(A) and A is saturated, p (V) is contained in
–1
-1
–1
A. It follows that both p (V) and q (V) equal all points of A that are mapped by p into V. To
check the second equation, we note that for any two subsets U and A of X, we have the inclusion
p(U A)  p(U) p(A)
To prove the reverse inclusion, suppose y = p(u) = p(a), for u  U and a  A. Since A is saturated,
–1
A contains the set p (p(a)), so that in particular A contains u. They y = p(u), where u  U A.
–1
Step (2): Now suppose A is open or p is open. Given the subset V of p(A), we assume that q (V)
is open in A and show that V is open in p(A).
–1
–1
Suppose first that A is open. Since q (V) is open in A and A is open in X, the set q (V) is open
–1
–1
in X. Since q (V) = p (V), the latter set is open in X, so that V is open in Y because p is a quotient
map. In particular, V is open in p(A).
–1
–1
–1
–1
Now suppose p is open. Since q (V) = p (V) and q (V) is open in A, we have p (V) = U A for
some set U open in X.
–1
Now p(p (V) = V because p is surjective, then
V = p(p (V)) = p(U A) = p(U) p(A)
–1
The set p(U) is open in Y because p is an open map; hence V is open in p(A).
Step (3): The proof when A or p is closed is obtained by replacing the word ‘open’ by the word
‘closed’ throughout step 2.
Theorem 2: Let p : X  Y be a quotient map. Let Z be a space and let g : X  Z be a map that is
–1
constant on each set p ({y}), for y  y. Then g induces a map f : Y  Z such that f o p = g. The
induced map f is continuous if and only if g is continuous; f is a quotient map if and only if g is
a quotient map.

LOVELY PROFESSIONAL UNIVERSITY 103