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This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".

For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.

The proof for the statement in bold is the following:

Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.

I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise.

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  • $\begingroup$ But you don't want to prove that $f(M - U_1 - U_2 - \cdots - U_n)$ is open, do you? Can you see why $U - F$ is open as soon as $U$ is open and $F$ closed? $\endgroup$ – PseudoNeo Sep 18 '14 at 9:20
  • $\begingroup$ @PseudoNeo- Sorry I meant what you wrote. $\endgroup$ – fierydemon Sep 18 '14 at 9:57
  • $\begingroup$ @PseudoNeo- Could you please give a proof of the fact that $F$ is closed? $\endgroup$ – fierydemon Sep 18 '14 at 9:58
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I'm going to assume that you by neighbourhood mean an open neighbourhood, else substitute that in the proof. Then observe that since $M$ is compact all closed subsets are compact. Thus since $N$ is hausdorff and f is continous the image of a closed set must be compact hence closed, thus you get $f(M-U_1-\dots-U_n)$ is closed. But an open set minus a closed set is open, since it just the intersection of the open set with the complement of the closed set, hence the intersection of two open sets.

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  • $\begingroup$ What do you mean by "image of a closed set must be continuous"? $\endgroup$ – fierydemon Sep 18 '14 at 9:58
  • $\begingroup$ But I think I got your answer. $\endgroup$ – fierydemon Sep 18 '14 at 10:01
  • $\begingroup$ Replace 'continuous' with 'compact'. $\endgroup$ – Moishe Kohan Sep 18 '14 at 10:27
  • $\begingroup$ Oh yes, I'm sorry about that it should of course have been: "the image of a closed set must be compact hence closed. $\endgroup$ – Adam Ehlers Nyholm Thomsen Sep 18 '14 at 10:41

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