S22 - 4th January 2005 Munkres ย 22 Ex 22.2(a The map p X...

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Unformatted text preview: 4th January 2005 Munkres ยง 22 Ex. 22.2. (a) . The map p : X โ†’ Y is continuous. Let U be a subspace of Y such that p- 1 ( U ) โŠ‚ 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 โ†’ Y is a quotient map. (b) . The map r : X โ†’ A is a quotient map by (a) because it has the inclusion map A , โ†’ X as right inverse. Ex. 22.3. Let g : R โ†’ A be the continuous map f ( x ) = x ร— 0. Then q โ—ฆ 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 ร— 1 x | x > } is closed but q ( A โˆฉ { x ร— 1 x | x > } = (0 , โˆž ) is not closed. The map q is not open for R ร— (- 1 , โˆž ) is open but q ( A โˆฉ ( R ร— (- 1 , โˆž ))) = [0 , โˆž ) is not open. Ex. 22.5. Let U โŠ‚ A be open in A . Since A is open, U is open in X . Since p is open, p ( U ) = q ( U ) โŠ‚ p ( A ) is open in Y and also in p ( A ) because p ( A ) is open [Lma 16.2]....
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S22 - 4th January 2005 Munkres ย 22 Ex 22.2(a The map p X...

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