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I was assigned the following homework problem for an introductory course in topology:

Let $p:\ X\to Y$ be a closed continuous surjective map such that $p^{-1}(\{y\})$ is compact for each $y\in Y$. Show that if $Y$ is compact, then $X$ is compact.

My question is, are the assumptions that $p$ be continuous and surjective even necessary for the result to hold? I've done the following:

Let $U$ be an open set containing $p^{-1}(\{y\})$. Since $X-U$ is closed, $p(X-U)$ is closed in $Y$. Then $W=Y-p(X-U)$ is an open set that contains $y$. Since $(X-U)\cap p^{-1}(W)=\varnothing$, $p^{-1}(\{y\})\subset p^{-1}(W) \subset U$.

Let $\{U_\alpha\}$ be an open cover of $X$. Then $p^{-1}(\{y\})\subset \bigcup_{k=1}^{N}U_k$. For each $y\in Y$, define $W_{y}$ as above; then $\{W_y\}$ is an open cover of $Y$, so can be covered by $W_{y_1},...,W_{y_m}$, and $p^{-1}(\{y_j\})\subset p^{-1}(W_{y_j})\subset \bigcup_{k=1}^{N}U_k$. Since $X=p^{-1}(Y)$, $X=\bigcup_{j=1}^{n} p^{-1}(W_{y_j})$. Since each $p^{-1}(W_{y_j})$ is covered by finitely many $U_k$, so is $X$, so it is compact.

If we dispense with surjectivity of $p$, the only thing that happens is that $p^{-1}(\{y\})$ may be empty, but then $p^{-1}(\{y\})\subset p^{-1}(W)$ holds vacuously, and the empty set is still compact.

If $p$ is continuous, then $p^{-1}(W)$ is open, but I don't use the fact that it is open at all in my proof.

I know that all 3 assumptions make a map "perfect" so that may be why they were mentioned, but can we dispense with continuity and surjectivity?

Thanks.

Stefan Hamcke
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KarlG
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  • Why don't you try thinking of it differently, not in the context of your problem? I.e., remove the assumption that $p$ is continuous and surjective, and try finding a counterexample to the homework problem statement. I don't know if it's right without the continuity and surjectivity condition, though. –  Aug 19 '14 at 00:09
  • If a map $p:X\to Y$ is continuous, closed, and has compact fibers, it is called a perfect map. Surjectivity isn't part of the definition, though some authors require it as well. Perfect maps are always proper, meaning that the preimage of a compact set is compact. If you show that for $B\subset Y$ the restriction $p':p^{-1}(B)\to B$ is perfect, you can use your proof here to show that $p$ is proper. – Stefan Hamcke Aug 19 '14 at 10:57
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    https://math.stackexchange.com/questions/1291151/prob-12-sec-26-in-munkres-topology-2nd-ed-how-to-show-that-the-domain-of-a-p – Saaqib Mahmood Sep 27 '18 at 06:44
  • @KarlG can you please also have a look at my post? Here is the link. https://math.stackexchange.com/questions/1291151/prob-12-sec-26-in-munkres-topology-2nd-ed-how-to-show-that-the-domain-of-a-p – Saaqib Mahmood Sep 27 '18 at 06:46
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    It has a problem when you assert that $( X - U ) \cap p^{-1}(W) = \emptyset$. Indeed, saying that $( X - U ) \cap p^{-1}(W) = \emptyset$ means that the whole preimage of $W$ is contained in $U$, which is not true in general. It may very well lie in $X - U$. For example, with $f : x \rightarrow x^2 $, we have $f^{-1} (f(1)) = {-1, 1}$. However, if $p$ is closed continuous surjective, then it is a quotient map. If you take $U$ a saturated open set containing $p^{-1}(y)$ then $p^{-1}(W) \subseteq U$. The rest of the proof then follows. The conditions "continuous" and "surjective" are thus nec – ntt Feb 09 '19 at 06:25
  • @ntt continuity and surjectivity are not needed. – Henno Brandsma May 14 '19 at 08:44

1 Answers1

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but can we dispense with continuity and surjectivity?

Yes. Your - correct - proof uses only the closedness of $p$ to show that every $y\in Y$ has an open neighbourhood $W_y$ such that $p^{-1}(W_y)$ is covered by finitely many of the $U_\alpha$, and since $Y$ is covered by finitely many $W_y$, it then follows that $X$ is covered by finitely many $U_\alpha$.

If one can prove something without using some assumptions in the statement of the theorem, these assumptions are then unnecessary to reach the conclusion.

Daniel Fischer
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  • Thank you, I was only worried that there was a hole in the proof, where I needed one of the other two assumptions to fill it. – KarlG Aug 19 '14 at 00:57
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    No hole, it's a nice proof. By the way, to see that the surjectivity isn't needed, note that by the closedness of $p$, the image $Z = p(X)$ is a closed, hence compact, subspace of $Y$, and considering the induced map $\tilde{p} \colon X\to Z$, we have a surjective closed [continuous, if we require that of $p$] map. So if the surjectivity were necessary, we could shift the target to $Z$, and by that we show that the surjectivity cannot be necessary. To see that the continuity cannot be necessary I have no such striking argument, unfortunately. – Daniel Fischer Aug 19 '14 at 01:04