Register to reply

Closed continuous surjective map and normal spaces

by radou
Tags: continuous, normal, spaces, surjective
Share this thread:
radou
#1
Nov20-10, 03:51 PM
HW Helper
radou's Avatar
P: 3,224
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.
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
micromass
#2
Nov21-10, 07:18 AM
Mentor
micromass's Avatar
P: 18,023
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].
micromass
#3
Nov21-10, 07:20 AM
Mentor
micromass's Avatar
P: 18,023
Also f(V) is not necessairily open...

radou
#4
Nov21-10, 02:31 PM
HW Helper
radou's Avatar
P: 3,224
Closed continuous surjective map and normal spaces

Quote Quote by micromass View Post
Also f(V) is not necessairily open...
Oh yes, I forgot the set we're looking for needs to be open! Hm, how could I make that right?
radou
#5
Nov21-10, 02:38 PM
HW Helper
radou's Avatar
P: 3,224
Actually, hold on, I just remembered there's a hint in the book - I'll think it through first.
micromass
#6
Nov21-10, 03:36 PM
Mentor
micromass's Avatar
P: 18,023
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...
radou
#7
Nov21-10, 04:46 PM
HW Helper
radou's Avatar
P: 3,224
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.
micromass
#8
Nov21-10, 04:53 PM
Mentor
micromass's Avatar
P: 18,023
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...
radou
#9
Nov21-10, 05:25 PM
HW Helper
radou's Avatar
P: 3,224
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.
micromass
#10
Nov21-10, 05:31 PM
Mentor
micromass's Avatar
P: 18,023
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.
radou
#11
Nov21-10, 05:37 PM
HW Helper
radou's Avatar
P: 3,224
Well W = Y \ p(X\U) contains all the images of the elements in U, right? And some elements must map to y.
micromass
#12
Nov21-10, 05:39 PM
Mentor
micromass's Avatar
P: 18,023
Yes!! It seems you've solved the problem then
radou
#13
Nov21-10, 05:44 PM
HW Helper
radou's Avatar
P: 3,224
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.
micromass
#14
Nov21-10, 05:48 PM
Mentor
micromass's Avatar
P: 18,023
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...
radou
#15
Nov21-10, 05:51 PM
HW Helper
radou's Avatar
P: 3,224
Quote Quote by micromass View Post
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...
Oh yes, the problem 12 on page 172 is a problem I skipped, and I see there's the same hint there.


Register to reply

Related Discussions
Spaces of continuous functions and Wronskians Linear & Abstract Algebra 7
Continuous functions in metric spaces Differential Geometry 1
Continuous expansion is surjective Calculus & Beyond Homework 2
Metric spaces and closed balls Calculus & Beyond Homework 7
Closed set, normed spaces Calculus & Beyond Homework 2