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According to Guillemin and Pollack, Differential Topology Page 109,

$f: X \to Y$ are appropriate for intersection theory ($X,Y$ are boundaryless oriented manifolds, $X$ is compact), when $Y$ is connected and has the same dimension as $X$, we define the degree of an arbitrary smooth map $f: X \to Y$ to be the intersection number of $f$ with any point $y$, $\deg(f) = I(f,\{y\})$.

Notice that in order to calculate $\deg(f)$, one simply selects any regular value $y$ and counts the preimage points $\{x:f(x) = y\}$, except that a point $x$ makes a contribution of $+1$ or $-1$ to the sum, depending on whether the isomorphism $df_x: T_x(X) \to T_y(Y)$ preserves or reverses orientation.

So I am confused here - what are the points making a contribution of $+1$ or $-1$ to the sum, and why we want to exclude them?

Thank you~

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  • $\begingroup$ I take $X$ compact here, else defining degree this way doesn't make sense! $\endgroup$ – Karthik C Jul 20 '13 at 6:58
  • $\begingroup$ Yeah, I'll add, thanks @AneeshKarthikC~:) $\endgroup$ – 1LiterTears Jul 20 '13 at 7:00
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They don't want to exclude. They are trying to say that you have to count the points with sign. That is, they take the inverse image. Enumerate it $\{x_1,x_2 ... x_n\}$

Here the author says that you can't count them and say that the degree is $n$.

I now elaborate what the author tries to say this way:

If at $df_{x_i}$ it preserves orientation, set $m_i=1$, else set $m_i=-1$

Now add the $m_i$s to get the degree.

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  • $\begingroup$ Aha, then that's all make sense!!!! Thank you so much Aneesh!!! I was so frustrated! Thanks! $\endgroup$ – 1LiterTears Jul 20 '13 at 7:08
  • $\begingroup$ Certainly, I just took a second to laugh and be relieved! $\endgroup$ – 1LiterTears Jul 20 '13 at 7:11

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