## Calculating the Degree of a Proper Map

As I asserted yesterday, there is a simple formula for the degree of a proper map . Pick any regular value of , some of which must exist, since the critical values have measure zero in . For each we define to be or as is orientation-preserving or orientation-reversing. Then I say that

The key is the inverse function theorem: the Jacobian of must have maximal rank at , so there’s some around on which is a diffeomorphism. This is true for each , so we get an open around each . Since is proper, there are only finitely many such , and they’re all separated from each other, so we can shrink each so that they’re all disjoint. Then we can set to be the intersection of all the , and we can further shrink each until its image is exactly .

That is, we’ve started with a regular value and its preimage , consisting of a bunch of points ; we’ve used the fact that is proper to widen it to an open set and its preimage, consisting of a bunch of disjoint open sets . Since is a diffeomorphism for each , is constantly orientation-preserving or orientation-reversing over all of , though of course each can be different.

Now if we pick some supported in , then is supported in the disjoint union of the . So we can calculate

Since the coefficient is independent of the choice of , this proves the assertion.

[...] 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 [...]

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