Closed continuous surjective map and normal spaces

[radou]

Homework Statement

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

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.

 

[micromass]

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 $$f(f^{-1}(A))=A$$.

 

[micromass]

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

 

[radou]

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]

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

 

[micromass]

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]

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]

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...

 

[radou]

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.

 

[micromass]

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.

 

[radou]

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

 

[micromass]

Yes! It seems you've solved the problem then

 

[radou]

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]

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]

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.