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.