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Unformatted text preview: 4th January 2005 Munkres 22 Ex. 22.2. (a) . The map p : X Y is continuous. Let U be a subspace of Y such that p 1 ( U ) X is open. Then f 1 ( p 1 ( U )) = ( pf ) 1 ( U ) = id 1 Y ( U ) = U is open because f is continuous. Thus p : X Y is a quotient map. (b) . The map r : X A is a quotient map by (a) because it has the inclusion map A , X as right inverse. Ex. 22.3. Let g : R A be the continuous map f ( x ) = x 0. Then q g is a quotient map, even a homeomorphism. If the compostion of two maps is quotient, then the last map is quotient; see [1]. Thus q is quotient. The map q is not closed for { x 1 x  x > } is closed but q ( A { x 1 x  x > } = (0 , ) is not closed. The map q is not open for R ( 1 , ) is open but q ( A ( R ( 1 , ))) = [0 , ) is not open. Ex. 22.5. Let U A be open in A . Since A is open, U is open in X . Since p is open, p ( U ) = q ( U ) p ( A ) is open in Y and also in p ( A ) because p ( A ) is open [Lma 16.2]....
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This note was uploaded on 01/12/2011 for the course MATH 110 taught by Professor Brown during the Fall '08 term at Arizona Western College.
 Fall '08
 Brown
 Topology

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