Calculate $\deg(f)$

Question

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~

I take $X$ compact here, else defining degree this way doesn't make sense! — Karthik C, Jul 20 '13 at 06:58

score 2 · Accepted Answer · answered Jul 20 '13 at 07:05

2

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.

answered Jul 20 '13 at 07:05

Karthik C

2,581

Aha, then that's all make sense!!!! Thank you so much Aneesh!!! I was so frustrated! Thanks! – 1LiterTears Jul 20 '13 at 07:08
Certainly, I just took a second to laugh and be relieved! – 1LiterTears Jul 20 '13 at 07:11

Calculate $\deg(f)$

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