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In Milnor's book "Topology from the Differentiable Viewpoint", He states the following lemma: Let $M, N$ be oriented $n$ manifolds, with $M$ compact and $N$ connected. Let $f : M \to N$ be a smooth map, and $y$ a regular value of $f$. Also, suppose $f$ can be extended to $F : X \to N$ where X is a compact oriented manifold with boundary M. Then $deg (f ; y) =0$.

I do not understand the proof given. He first proves it in the case when $y$ is a regular value of $F$. In this case, he claims that $F^{-1} (y) $ is a compact $1$ manifold with boundary and hence is the disjoint union of circles and line segments. He then claims that for any line segment, the sum of signs of the two boundary points is $0$ establishing the theorem. He then extends it to the other points by using the fact that it (the degree) is locally constant. I don't understand Milnor's proof of the claim.

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  • $\begingroup$ Actually, judging by what you have written, it seems that you understand the proof perfectly. Could you specify which part exactly it is that you are having difficulties with? $\endgroup$ – Amitai Yuval Jun 4 '15 at 14:28
  • $\begingroup$ I don't understand the proof of the claim that the two opposite points have opposite sign. $\endgroup$ – Ishan Banerjee Jun 4 '15 at 14:29
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Let $c\subset f^{-1}(y)$ be a segment, and let $p,q$ denote the two endpoints of $c$. Note that $p,q\in\partial X=M$. We equip $c$ with the orientation induced by $F$ and the orientations of $X$ and $N$. With no loss of generality, this orientation points inwards at $p$ and outwards at $q$. It follows that at $df_q$ carries an oriented basis of $T_qM$ to an oriented basis of $T_{f(q)}N$, whereas $d_pf$ carries an oriented basis of $T_pM$ to a non-oriented basis of $T_{f(p)}N$.

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  • $\begingroup$ What exactly is the orientation induced by F and the orientations of X and N? $\endgroup$ – Ishan Banerjee Jun 4 '15 at 15:25
  • $\begingroup$ @IshanBanerjee It is explained in Milnor's book, in the middle of the proof you're asking about (at least in the edition I have). $\endgroup$ – Amitai Yuval Jun 5 '15 at 6:22
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Was wondering about the exact same thing. Found an answer here: http://web.stanford.edu/~amwright/PoincareHopf.pdf Page 3 Lemma 3

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  • $\begingroup$ you mean page 2? $\endgroup$ – Kamil Jarosz Dec 16 '15 at 19:56

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