The Unapologetic Mathematician

Mathematics for the interested outsider

Calculating the Degree of a Proper Map

As I asserted yesterday, there is a simple formula for the degree of a proper map f:M\to N. Pick any regular value q\in N of f, some of which must exist, since the critical values have measure zero in N. For each p\in f^{-1}(q) we define \mathrm{sgn}_pf to be +1 or -1 as f is orientation-preserving or orientation-reversing. Then I say that

\displaystyle\deg f=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf

The key is the inverse function theorem: the Jacobian of f must have maximal rank at p, so there’s some U\subseteq M around p on which f:U\to f(U)\subseteq N is a diffeomorphism. This is true for each p\in f^{-1}(q), so we get an open U_p around each p. Since f is proper, there are only finitely many such p, and they’re all separated from each other, so we can shrink each U_p so that they’re all disjoint. Then we can set V\subseteq N to be the intersection of all the f(U_p), and we can further shrink each U_p until its image is exactly V.

That is, we’ve started with a regular value q\in N and its preimage f^{-1}(q), consisting of a bunch of points p\in M; we’ve used the fact that f is proper to widen it to an open set V and its preimage, consisting of a bunch of disjoint open sets U_p\subseteq M. Since f:U_p\to V is a diffeomorphism for each p\in f^{-1}(q), f is constantly orientation-preserving or orientation-reversing over all of U_p, though of course each U_p can be different.

Now if we pick some \omega\in\Omega_c^n(N) supported in V, then f^*\omega is supported in the disjoint union of the U_p. So we can calculate

\displaystyle\begin{aligned}\int\limits_Mf^*\omega&=\int\limits_{\biguplus_{p\in f^{-1}(q)}U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\int\limits_{U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\int\limits_{f(U_p)}\omega\\&=\left(\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\right)\int\limits_V\omega\\&=\deg f\int\limits_V\omega\end{aligned}

Since the coefficient \deg f is independent of the choice of \omega, this proves the assertion.

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December 10, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. [...] what does this mean? The identity map has degree , while we can calculate that the antipodal map has degree . Since these are different, the two maps must not act [...]

    Pingback by The “Hairy Ball Theorem” « The Unapologetic Mathematician | December 13, 2011 | Reply

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