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Math Help - [SOLVED] Algebraic Topology: Covering Spaces

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    [SOLVED] Algebraic Topology: Covering Spaces

    Let p: \overset{\sim}{X} --> X be a covering and let f, g: Y --> \overset{\sim}{X} be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JJMC89 View Post
    Let p:X~ --> X be a covering and let f, g: Y --> X~ be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.
    I am just a little confused. Usually the notation X/\sim is used to denote a quotient space (by some equivalence relation \sim) but I think you might just mean \overset{\sim}{X}.

    Ok, so I have a visual here is what we have


    Right?
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  3. #3
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    Correct.
    I edited my post to show that it is \overset{\sim}{X}.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JJMC89 View Post
    Correct.
    I edited my post to show that it is \overset{\sim}{X}.
    Hmm...I thought I had it but I don't. Let me try to think of something else. A common trick when doing things like this is to consider that map f\oplus g:Y\to\overset{\sim}{X}\times\overset{\sim}{X}:x\m  apsto(f(x),g(x)).

    Hmm, obviously the tricks going to have to do something with if x\in A(f,g)=\left\{y\in Y:f(y)=g(y)\right\} then the fibers of p(f(x)),p(g(x)) are going to be same.

    I'm sorry for getting your hopes, I'll try to figure it out.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry for getting your hopes, I'll try to figure it out.
    Thanks
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  6. #6
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    Quote Originally Posted by JJMC89 View Post
    Let p: \overset{\sim}{X} --> X be a covering and let f, g: Y --> \overset{\sim}{X} be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.
    Let K=\{y \in Y| f(y) = g(y)\} and let y_0 be an arbitrary point of the closure of K. Let x_0 = pf(y_0)=pg(y_0). We shall show that y_0 \in K, which implies that K is a closed set of Y.
    Suppose to the contrary that y_0 \notin K. Then f(y_0) \neq g(y_0). Let U be an open neighborhood of x_0. Then p^{-1}(U) contains f(y_0) and g(y_0), respectively. Let V and W be open components of p^{-1}(U) such that f(y_0) \in V and g(y_0) \in W, where V and W are disjoint open sets. By hypothesis, f and g are continuous, so there exist open neighborhoods M of y_0 such that f(M) \subset V and g(M) \subset W. This contradicts that y_0 is a point of the closure of K. Thus y_0 \in K and K is a closed set of Y.

    To show that K is open in Y, you need to show that for any y_0 \in K, there exists an open set contains y_0 and is contained in K. Let V and W be an open component of p^{-1}(U) such that f(y_0) \in V and g(y_0) \in W, where U is an open neighborhood of x_0=pf(y_0)=pg(y_0). Since y_0 \in K, we have V=W. Thus K is an open set of Y.
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