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Let $p \colon X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1}\big(\{ y \} \big)$ is compact for each $y \in Y$. (Such a map is called a perfect map.) (a) Show that if $X$ is Hausdorff, then so is $Y$.

My attempt:

Approach(1): Let $x,y \in Y$ with $x\neq y$. So $p^{-1}(\{x\}),p^{-1}(\{y\})$ is compact and $p^{-1}(\{x\}) \cap p^{-1}(\{y\})= p^{-1}(\{x\} \cap \{y\})$$= p^{-1}(\emptyset)=\emptyset$. By exercise 5 section 26, $\exists U,V\in \mathcal{T}_X$ such that $p^{-1}(\{x\}) \subseteq U$, $p^{-1}(\{y\})\subseteq V$ and $U\cap V=\emptyset$. By generalize hint/chapter 3 theorem 11.2 of Dugundji topology, $\exists R \in \mathcal{N}_x$ such that $p^{-1}(R)\subseteq U$ and $\exists S\in \mathcal{N}_y$ such that $p^{-1}(S)\subseteq V$. Since $U\cap V=\emptyset$, we have $p^{-1}(R) \cap p^{-1}(S)=\emptyset =p^{-1}(R\cap S)$. Since $p$ is surjective, $R\cap S=\emptyset$. Hence $\exists R,S\in \mathcal{T}_Y$ such that $x\in R$, $y\in S$ and $R\cap S=\emptyset$. Is this proof correct?

Approach(2): It’s easy to check, normal $\Rightarrow$ regular $ \Rightarrow$ $T_2$ $\Rightarrow$ $T_1$. By Exercise 6, Section 31 of Munkres’ Topology, $Y$ is $T_2$. Is this proof correct?

Note: In approach(1) we used $p^{-1}(y)$ is compact, $p$ is closed map and $p$ is surjective conditions. In approach(2) we used $p$ is continuous, closed and surjective conditions.

user264745
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1 Answers1

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Edit: I am expanding this instead of commenting since stack exchange doesn’t seem to like so many comments:

Approach 2 does not work. In order to use your cited theorem, you would need the following 1 $X$ is normal. 2 a continuous, closed, surjective function from $X$ to $Y$

You have the following:

1 $X$ is Hausdorff. 2 a continuous, closed, surjective function from $X$ to $Y$ that has this other strong property.

So, to use the first theorem, you would need to show that $X$ in your new theorem is regular. That is not the case. The original theorem is weaker then if you replace normal with Hausdorff. Meaning it applies to much fewer situations, specifically it does not apply to situations with $X$ is not normal as we have.

Admittedly, I would save us both some headache if I found an example a a Hausdorff space that is not regular, and a closed continuous surjective map that is not perfect such that $Y$ is not Hausdorff. Unfortunately that is probably a pretty nasty example and is beyond me at the moment.

For approach 1, assuming you applied each theorem correctly this looks right to me. My main question is that the “generalize hint” talks about maps between compact spaces, do you know why it works for Hausdorff spaces? Assuming the other theorems are used correctly this looks fine to me.

Nico Tripeny
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  • Normal property is stronger than Hausdorff, no? For generalize hint, you misinterpreted something, I guess. – user264745 May 17 '22 at 18:08
  • Exactly. You have that $X$ is Hausdorff so is in general not Normal. The theorem you cite uses that $X$ is normal, so does not apply here. – Nico Tripeny May 17 '22 at 18:12
  • A normal space is $T_2$. We don’t need equivalent. – user264745 May 17 '22 at 18:13
  • Yes it’s even $T_4$. My point is $X$ in your question is $T_2$ not $T_4$ right? So you can not use theorems about Normal spaces to help you. – Nico Tripeny May 17 '22 at 18:16
  • Either you or me making basic mistake. As far as I know, this problem is special case of previous problem(exercise 6 section 31). – user264745 May 17 '22 at 18:19
  • I agree. I still think I’m correct but am trying to be cautious. The previous problem says basically: assume $X$ is normal with a closed continuous surjective map. This problem is: assume $X$ is Hausdorff (seeker) with a closed continuous surjective map such that $f^{-1}(p)$ is compact (stronger). So I don’t see how this is a special case. It changes weakness and strong ness in different ways in different places – Nico Tripeny May 17 '22 at 18:22
  • In previous problem we didn’t use entire conditions of perfect map, precisely $p^{-1}(y)$ is compact condition. Read note part in my post. A normal space is automatically $T_2$, agree? – user264745 May 17 '22 at 18:27
  • Do you wanna bet rep? – user264745 May 17 '22 at 18:44
  • Unfortunately I have very little (perhaps fair you don’t trust me). But I agree a normal space is Hausdorff. Do you agree that $X$ in the question is not normal? – Nico Tripeny May 17 '22 at 18:46
  • Of course I agree $X$ in this problem is not normal. I think, we have to wait for verdict from some independent individual. – user264745 May 17 '22 at 18:50
  • Perhaps I found a mistake for both of us that could be satisfying? In fact, normal and $T_4$ are not the same, normal plus Hausdorff is $T_4$ but normal does not imply Hausdorff. So either way for the earlier conversation, I think this would imply the other lemma cannot be used. Does that seem right? Admittedly Munkre’s might use a different notation. – Nico Tripeny May 17 '22 at 19:56
  • You’re using wrong definition of normal. Normal is $T_4$ and $T_1$. – user264745 May 17 '22 at 20:00
  • I see. Normal is defined differently elsewhere, not necessarily $T_1$ – Nico Tripeny May 17 '22 at 20:05
  • But you’re using something entirely different. – user264745 May 17 '22 at 20:06
  • As I said, my comment two times ago was incorrect as I was using a different definition. I stand by the rest of my comments. – Nico Tripeny May 17 '22 at 20:07
  • You wrote normal plus Hausdorff($T_2$) is $T_4$, that is not different, that’s wrong. – user264745 May 17 '22 at 20:09
  • Normal and Hausdorff is $T_4$. That is not the definition and I did not mean it as such. But certainly a space that is both normal and Hausdorff is $T_4$ using either of our definitions correct? To clarify, my definition was that disjoint closed sets are separable, Munkres is that on top of single points are closed. – Nico Tripeny May 17 '22 at 20:10
  • See https://math.stackexchange.com/q/4457432/861687. I was wrong. – user264745 May 25 '22 at 12:18