## The Degree of a Map

This will be a nice, simple definition. As we saw last time, the top-degree compactly-supported de Rham cohomology is easy to calculate for a connected, oriented manifold . Specifically, we know that .

So, let’s say that is a smooth map from one connected, oriented -manifold to another. The pullback induces a map on cohomology: . But since each of these cohomology spaces is (isomorphic to) , must act on them by multiplication by some number. That is, if is a closed -form, and if the integral of over is , then is also a closed -form, and the integral of over is , where is some real number that depends only on the function and not on the form . That is, the same works for all .

We call this number the “degree” of , and write . It turns out that when is a proper map — one for which the preimage of any compact set is compact, which is always the case when is itself compact — that we have a good way of calculating it. If is a regular value of , some of which which will always exist, then the preimage of consists of a finite collection of points . For each one we calculate the “signum” of at — written — to be if is orientation-preserving and if is orientation-reversing. I say that

While we will not prove this until next time, there is one immediate consequence: if is not onto, then any point in has an empty preimage, and so we find that . In a sense, the degree of is counting how many times the image of covers that of , considering orientation.

Are you sure “f” doesn’t just need more cowbell?

Comment by Rhonda ROO | December 9, 2011 |

[…] 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 […]

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[…] can use the concept of degree to prove the (in)famous “hairy ball theorem”. That is, any smooth vector field defined […]

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