proper differentiable maps of manifolds 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. 
 A: 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)
A: 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)   
