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

    Then he gives the following example:

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

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

    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" $\displaystyle [0,1]$ of $\displaystyle X$ is not open in $\displaystyle Y$.

    Can someone please tell me:
    (i) how to verify that the map is a closed map
    (ii) how $\displaystyle [0,1]$ can be considered open in $\displaystyle X$
    (iii) how then is image of $\displaystyle [0,1]$ not open in $\displaystyle Y$
    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 $\displaystyle {X-[0,1]}$ is $\displaystyle [2,3]$, which is clearly closed $\displaystyle \longrightarrow[0,1]$ is open in $\displaystyle X$, but what about (i) and (iii)?
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  3. #3
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    Sorry for the bother, I got it:

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

    (iii) $\displaystyle p([0,1])=[0,1]$ and $\displaystyle (Y-[0,1])=]1,2]$, which is not closed; thus, $\displaystyle p([0,1])$ is not open in $\displaystyle Y$, and so it follows that $\displaystyle p:X \rightarrow Y$ 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 $\displaystyle Y = [0,2]$? Is $\displaystyle [0,1]$ 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 $\displaystyle Y = [0,2]$? Is $\displaystyle [0,1]$ open in Y?
    $\displaystyle X=[0,1]\cup[2,3] \subset \mathbb{R}$
    $\displaystyle Y=[0,2] \subset \mathbb{R}$

    $\displaystyle p$ is a map from $\displaystyle X$ to $\displaystyle Y$.

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

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

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