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Closed continuous surjective map and normal spaces 
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#1
Nov2010, 03:51 PM

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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 welldefined preimage, right? Thanks in advance. 


#2
Nov2110, 07:18 AM

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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
Nov2110, 02:31 PM

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Closed continuous surjective map and normal spaces



#5
Nov2110, 02:38 PM

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Actually, hold on, I just remembered there's a hint in the book  I'll think it through first.



#6
Nov2110, 03:36 PM

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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... 


#7
Nov2110, 04:46 PM

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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.



#8
Nov2110, 04:53 PM

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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... 


#9
Nov2110, 05:25 PM

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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. 


#10
Nov2110, 05:31 PM

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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.



#11
Nov2110, 05:37 PM

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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
Nov2110, 05:44 PM

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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. 


#14
Nov2110, 05:48 PM

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Yes, you will need the hint on problem 7 to and the hint is also used in problem 12, page 172. It's a neat trick and worth knowing about...



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