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Math Help - topology, continuity

  1. #1
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    topology, continuity

    Prove the following are equivalent:
    (a) \forall V \subseteq Y, V is an open subset of Y iff p^{-1}(V) is an open subset of X
    (b) F a closed subset of Y iff p^{-1}(F) is a closed subset of X.
    Last edited by zelda2139; January 16th 2009 at 06:48 AM.
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  2. #2
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    Hello,

    is stronger than continuity.
    what do you mean by this ?
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  3. #3
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    I am not sure, we care covering strong continuity, but we didn't go over it a lot, that's why I'm stuck on this one.
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  4. #4
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    I am not 100% sure but I think the question (a) asks to show an "quotient map" is stronger than "continuous map".

    In (a) "iff" corresponds an "quotient map" and "if" corresponds a "continuous map".
    Last edited by aliceinwonderland; May 13th 2009 at 10:19 PM. Reason: Even though the original question was gone, I accidently found that this needs a correction (open map -> quotient map)
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    I figured it out.
    Last edited by zelda2139; January 16th 2009 at 06:50 AM.
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    I will do one way for you if you promise to show us all the other way.
    a \Rightarrow b
    If F \subseteq Y\,\& \,F is closed then Y\backslash F is open in Y.
    Therefore, p^{ - 1} \left( {Y\backslash F} \right) is open in X.
    But p^{ - 1} \left( {Y\backslash F} \right) = p^{ - 1} \left( Y \right)\backslash p^{ - 1} \left( F \right) = X\backslash p^{ - 1} \left( F \right) is open.
    Therefore, p^{ - 1} \left( F \right) is closed in X.
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  7. #7
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    b \Rightarrow a, let V be closed in Y. Y - V is closed in Y, so that p^{ - 1}(Y - V) is closed in X. Thus X - p^{ - 1}(Y - V) = X - (p^{ - 1}(Y) - p^{ - 1}(V)) = X - (X - p^{ - 1}(V)) = p^{ - 1}(V) is open in X. Every statement made works when starting with p^{ - 1}(V) open in X. Q.E.D.
    Last edited by zelda2139; January 16th 2009 at 06:50 AM.
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