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Math Help - Topology Quotient Space

  1. #1
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    Topology Quotient Space

    In Topology 2ed by Munkres, the following definition is given of a quotient map:

    Let X,Y be topological spaces; let p:{X}\rightarrow{Y} be a surjective map. The map p is said to be a quotient map provided a subset {U}\subset{Y} is open in Y if and only if {p^{-1}}(U) is open in X.

    Then he gives the following example:

    Let X be the subspace [0,1]\cup[2,3] of \mathbb{R}. The map p:{X}\rightarrow{Y} defined by:

    p(x)=x if  x\in [0,1]
    p(x)={(x-1)} if  x\in [2,3]

    In order to show that this map is a quotient map, one needs to show that is is surjective, continuous and closed. The first two are clear, however I am not so sure about how to verify that it is a closed map.

    Also, he notes that the map is not open, saying as proof that the image of the "open set" [0,1] of X is not open in Y.

    Can someone please tell me:
    (i) how to verify that the map is a closed map
    (ii) how [0,1] can be considered open in X
    (iii) how then is image of [0,1] not open in Y
    Last edited by TXGirl; May 17th 2008 at 02:04 PM.
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  2. #2
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    I guess for (ii), I simply say that {X-[0,1]} is [2,3], which is clearly closed \longrightarrow[0,1] is open in X, but what about (i) and (iii)?
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  3. #3
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    Sorry for the bother, I got it:

    (i) p([0,1])=[0,1] is closed in Y and p([2,3])=[1,2] is also closed in Y thus p:X \rightarrow Y is a closed map

    (iii) p([0,1])=[0,1] and (Y-[0,1])=]1,2], which is not closed; thus, p([0,1]) is not open in Y, and so it follows that p:X \rightarrow Y is not an open map
    Last edited by TXGirl; May 17th 2008 at 01:44 PM.
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  4. #4
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    What is Y? Is it Y = [0,2]? Is [0,1] open in Y?
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  5. #5
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    Quote Originally Posted by Plato View Post
    What is Y? Is it Y = [0,2]? Is [0,1] open in Y?
    X=[0,1]\cup[2,3] \subset \mathbb{R}
    Y=[0,2] \subset \mathbb{R}

    p is a map from X to Y.

    [0,1] is open in X because its complement in X is [2,3], which is closed.

    p([0,1]) is not open in Y because its complement in Y is ]1,2], which is not closed.

    The complement of [0,1] in Y is ]1,2] , which is not closed, so no, [0,1] is not open in Y
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