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Closed continuous surjective map and Hausdorff space

  1. Nov 22, 2010 #1

    radou

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    1. The problem statement, all variables and given/known data

    Here's a nice one. I hope it's correct.

    Let p : X --> Y be a closed, continuous and surjective map such that p^-1({y}) is compact for every y in Y. If X is Hausdorff, so is Y.

    3. The attempt at a solution

    Let y1 and y2 in Y. p^-1({y1}) are then p^-1({y2}) disjoint and compact subsets of X. Since X is Hausdorff, for p^-1({y1}) and for any x in p^-1({y2}) there exist disjoint open sets U and V containing p^-1({y1}) and x, respectively. Now find such pair of open sets for p^-1({y1}) and for any x in p^-1({y2}). These sets form open covers for p^-1({y1}) and for p^-1({y2}) respectively, so they have finite subcovers. Take the intersection of all sets from the finite subcover for p^-1({y1}), let's call it U1. Take the union of all sets from the finite subcover for p^-1({y2}), call it U2. U1 and U2 are disjoint.

    Now, since U1 and U2 are open sets containing p^-1({y1}) and p^-1({y2}) respectively, there exist neighborhoods W1 of y1 and W2 of y2 such that p^-1(W1) is contained in U1 and p^-1(W2) is contained in U2.

    I claim that W1 and W2 are disjoint.

    Suppose they were not - let y be an element in their intersection. Then p^-1({y}) is contained both in U1 and U2, contradicting the fact that U1 and U2 are disjoint.
     
  2. jcsd
  3. Nov 22, 2010 #2

    micromass

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    Your proof is perfect!
     
  4. Nov 22, 2010 #3

    radou

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    Excellent! This tradition mustn't go on, since I'll start to think I'm good :D

    Btw, the "hint" is extremely useful. I don't see another way we could "generate" an open set with certain required properties in the codomain.
     
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