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Closed continuous surjective map and normal spaces

  1. Nov 20, 2010 #1

    radou

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

    Let p : X --> Y be a closed, continuous and surjective map. Show that if X is normal, so is Y.

    3. The attempt at a solution

    I used the following lemma:

    X is normal iff given a closed set A and open set U containing A, there is an open set V containing A and whose closure is contained in U.

    So, let A be a closed set in Y, and U some neighborhood of A. By continuity, f^-1(A) is closed in X, and f^-1(U) is open in X. Further on, f^-1(U) is an open neighborhood of f^-1(A), since it contains f^-1(A). If we apply the lemma above to these sets, we can find an open set V which satisfies the criterion, since X is regular. Since p is a closed map, the image of Cl(V) is closed, and since all the inclusions remain preserved, Y is normal.

    By the way, just to check, the requirement for p to be surjective was because we have chosen a set in Y, and surjectivity guarantees that this set has a well-defined preimage, right?

    Thanks in advance.
     
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  3. Nov 21, 2010 #2

    micromass

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    Yes, the idea is correct. But you did use surjectivity in an essential way. Let me point out where: you have the following situation in X:

    [tex]f^{-1}(A)\subseteq V\subseteq Cl(V)\subseteq f^{-1}(U)[/tex]

    By taking the image of f, you obtain


    [tex]f(f^{-1}(A))\subseteq f(Cl(V))\subseteq f(f^{-1}(U))[/tex]

    This is not want you want... You want

    [tex]A\subseteq f(Cl(V))\subseteq U[/tex]

    This is where surjectivity comes in, since then it holds that [tex]f(f^{-1}(A))=A[/tex].
     
  4. Nov 21, 2010 #3

    micromass

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    Also f(V) is not necessairily open...
     
  5. Nov 21, 2010 #4

    radou

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    Oh yes, I forgot the set we're looking for needs to be open! Hm, how could I make that right?
     
  6. Nov 21, 2010 #5

    radou

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    Actually, hold on, I just remembered there's a hint in the book - I'll think it through first.
     
  7. Nov 21, 2010 #6

    micromass

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    Ah yes, I can see there's a hint. I don't think it is possible to prove this without that hint...

    What that hint asks you to prove, is actually equivalent to closedness of maps...
     
  8. Nov 21, 2010 #7

    radou

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    I think I see how I can prove it with using the hint, but I have problems proving the hint :) I'll continue thinking about it, but it's starting to drive me crazy.
     
  9. Nov 21, 2010 #8

    micromass

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    You should only use closedness for proving the hint. I'll get you started:

    Take U open such that [tex]p^{-1}(y)\subseteq U[/tex]. Then [tex]X\setminus U[/tex] is closed. The closedness of p yields that [tex]p(X\setminus U)[/tex] is closed. Thus [tex]W:=Y\setminus p(X\setminus U)[/tex] is open. Now show that this W satisfies all our desires...
     
  10. Nov 21, 2010 #9

    radou

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    p^-1(W) = p^-1(Y) \ p^-1(p(X\U)) = X \ p^-1(p(X\U)), and since X\U[tex]\subseteq[/tex]p^-1(p(X\U)), p^-1(W) is contained in U, right?

    Now, apply out hint to the set V, which is open and contains p^-1({a}), for any a in A, we can find a neighborhood Wa of a such that p^-1(Wa) is contained in V.

    Now, any Wa is contained in p(V), so the union W of all the Wa's along the set A is contained in p(V), right? And hence, in p(Cl(V)), too. SO, the closure of W is contained in p(CL(V)), so W and its closure are the sets we needed to find.

    Uhh, I feel this is very slippery.
     
  11. Nov 21, 2010 #10

    micromass

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    Yes, that seems all correct. There's just one thing: you didn't prove the hint completely: you still need to show that W is a neighbourhood of y (thus [tex]y\in W[/tex]), but this shouldn't give to much of a problem.
     
  12. Nov 21, 2010 #11

    radou

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    Well W = Y \ p(X\U) contains all the images of the elements in U, right? And some elements must map to y.
     
  13. Nov 21, 2010 #12

    micromass

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    Yes!! It seems you've solved the problem then :smile:
     
  14. Nov 21, 2010 #13

    radou

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    Excellent! Thanks!

    Btw, I have a feeling I'll be needing that "hint" on some other problems :) It's interesting how "at a first glance moderately easy" problems can cause a respectable amount of trouble.
     
  15. Nov 21, 2010 #14

    micromass

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    Yes, you will need the hint on problem 7 to :smile: and the hint is also used in problem 12, page 172. It's a neat trick and worth knowing about...
     
  16. Nov 21, 2010 #15

    radou

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    Oh yes, the problem 12 on page 172 is a problem I skipped, and I see there's the same hint there.
     
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