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Is the following statement true?If yes, why?

Let $f: M\to N$ be a proper morphism between smooth manifolds. Let $x$ be a point of $N$, and $U$ a nbhd of $f^{-1}(x)$ in $M$. Then there exists a nbhd $V$ of $x$ in $N$ such that $f^{-1}(V)\subset U$.

Here, proper means that the preimage of any compact set is compact.

It seems to me this was used in a expository article that I am reading. If it is true, I expect it to be true for proper morphism of locally compact topological spaces. But for some reason I wasn't able to find a proof.

Thank you.

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After reading elgeorges's answer, it occurs to me that the following two properties are in fact equivalent (where $X, Y$ are topological spaces and $f$ is continuous): (1) $f:X\to Y$ is closed. (2) For any point $y\in Y$ and an open set $U\supset f^{-1}(y)$ in $X$, there exists an open nbhd $V$ of $y$ such that $f^{-1}(V)\subseteq U$. – Jiangwei Xue May 13 '11 at 14:19
Absolutely correct, Jiangwei: I had never realized we had an equivalence, so thanks for pointing it out. – Georges Elencwajg May 13 '11 at 15:30
up vote 4 down vote accepted

Suppose not. Then there is a sequence $(y_n)_{n\geq1}$ in $M\setminus U$ such that $f(y_n)\to x$. The set $S=\{f(y_n):n\geq1\}\cup\{x\}$ is compact, so its preimage $f^{-1}(S)$ is also compact. Since the sequence $(y_n)_{n\geq1}$ is contained in $f^{-1}(S)$, we can —by replacing it with one of its subsequences, if needed— assume that in fact $(y_n)_{n\geq1}$ converges to a point $y\in M$.

Can you finish?

(I am using that $x$ has a countable basis of neighborhoods here and that sequences in a compact subset of $M$ have convergent subsequences —to reduce to dealing with sequences— but more technology will remove that in order to generalize this to spaces other that manifolds)

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Perhaps $S = \{ f(y_n\}: n \geq 1\} \cup \{ x \}$ ? – Raziel May 13 '11 at 7:50
Yes, I can. Thank you very much. – Jiangwei Xue May 13 '11 at 8:14

You are quite correct that this is a purely topological statement. Proper maps are closed and here is a proof that works for closed maps between topological spaces, not necessarily manifolds nor even locally compact .
Proof: Since $U$ is open in $M$, $M \setminus U$ is closed, hence (by the assumption "closed map" ) so is $F\subset N$ defined by $F=f(M\setminus U) $. Just take $V=N\setminus F$. Done.

Edit: Since the issue of "properness" is a very interesting one, I hope you will excuse the following development, prompted by Theo's comment.
The principal property of proper maps is that they are closed and I think this should be built into the definition (obviously Algebraic Geometry makes its influence felt here). So a nice definition is that a continuous map $f:X\to Y$ between topological spaces is proper if it is closed and if the inverse images of points in $y \in Y$ are compact subsets $f^{-1}(y) \subset X$ [technically some say "quasi-compact" subsets to stress that no Hausdorff condition is assumed]. The link with the usual notion of "properness" is then the following theorem (in which proper has the meaning just given), whose assumptions are pleasantly weaker than one might expect. The point I advocate is that one should learn it once and for all, and that then you can often give short elegant proofs in the context of properness.

Theorem If $X$ is Hausdorff and $ Y $ locally compact, then $f$ is proper if and only if the inverse image of a compact subset $K\subset Y$ is a compact subset $f^{-1}(K) \subset Y$.

(The proof is in Chapter 1 of Bourbaki's book on General Topology, which adopts this point of view)

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This argument is very nice. But in showing that proper maps are closed I think you need to use an argument similar to Mariano's. Do I miss something? – t.b. May 13 '11 at 8:23
Ah, Theo, this is the heart of the matter. Yes, you have to prove something if you want to show that "proper" maps are closed. But the optimal statement is weaker than one might think: only the target has to be locally compact. And, much more importantly, you should do that once in your life and then freely use it.My experience is that most of the time, when people want to use properness they end up actually proving closedness (more or less) and this obscures the issue at hand.It is as if, inside a proof of a property of a compact subset, you had to reprove that it is closed.(To be cont'd) – Georges Elencwajg May 13 '11 at 8:51
Since, in my experience, the issue comes up quite often (for example in the theory of covering spaces), I'll add an edit to my question , which will tell you why I put "proper" in quotes in my comment above. – Georges Elencwajg May 13 '11 at 8:58
Thanks for your elaborations. I definitely agree with all the things you are saying and I don't have a problem with your argument at all! I just wanted to point out that closedness of proper maps is a useful and important feature. Actually, if $X$ is Hausdorff, $Y$ is locally compact Hausdorff and $f: X \to Y$ is proper in the sense that pre-images of compact sets are compact then $X$ is automatically locally compact, but maybe that's what you call the optimal statement. – t.b. May 13 '11 at 9:03
Theo, coming back from my editing, I'm happily surprised by your new comment: I couldn't agree more with it! Thanks for your interest. – Georges Elencwajg May 13 '11 at 9:48

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