Closed continuous surjective map and normal spaces

1. Nov 20, 2010

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?

2. Nov 21, 2010

micromass

Staff Emeritus
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:

$$f^{-1}(A)\subseteq V\subseteq Cl(V)\subseteq f^{-1}(U)$$

By taking the image of f, you obtain

$$f(f^{-1}(A))\subseteq f(Cl(V))\subseteq f(f^{-1}(U))$$

This is not want you want... You want

$$A\subseteq f(Cl(V))\subseteq U$$

This is where surjectivity comes in, since then it holds that $$f(f^{-1}(A))=A$$.

3. Nov 21, 2010

micromass

Staff Emeritus
Also f(V) is not necessairily open...

4. Nov 21, 2010

Oh yes, I forgot the set we're looking for needs to be open! Hm, how could I make that right?

5. Nov 21, 2010

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

6. Nov 21, 2010

micromass

Staff Emeritus
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. Nov 21, 2010

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. Nov 21, 2010

micromass

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

Take U open such that $$p^{-1}(y)\subseteq U$$. Then $$X\setminus U$$ is closed. The closedness of p yields that $$p(X\setminus U)$$ is closed. Thus $$W:=Y\setminus p(X\setminus U)$$ is open. Now show that this W satisfies all our desires...

9. Nov 21, 2010

p^-1(W) = p^-1(Y) \ p^-1(p(X\U)) = X \ p^-1(p(X\U)), and since X\U$$\subseteq$$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. Nov 21, 2010

micromass

Staff Emeritus
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 $$y\in W$$), but this shouldn't give to much of a problem.

11. Nov 21, 2010

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

12. Nov 21, 2010

micromass

Staff Emeritus
Yes!! It seems you've solved the problem then

13. Nov 21, 2010

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. Nov 21, 2010

micromass

Staff Emeritus
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...

15. Nov 21, 2010