The Unapologetic Mathematician

Mathematics for the interested outsider

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 M. Specifically, we know that H_c^n(M)\cong\mathbb{R}.

So, let’s say that f:M\to N is a smooth map from one connected, oriented n-manifold to another. The pullback f^* induces a map on cohomology: f^*:H_c^n(N)\to H_c^n(M). But since each of these cohomology spaces is (isomorphic to) \mathbb{R}, f^* must act on them by multiplication by some number. That is, if \omega\in\Omega_c^n(N) is a closed n-form, and if the integral of \omega over N is I, then f^*\omega\in\Omega_c^n(M) is also a closed n-form, and the integral of f^*\omega over M is d_fI, where d_f is some real number that depends only on the function f and not on the form \omega. That is, the same d_f works for all \omega.

We call this number the “degree” of f, and write \deg f. It turns out that when f is a proper map — one for which the preimage of any compact set is compact, which is always the case when M is itself compact — that we have a good way of calculating it. If q\in N is a regular value of f, some of which which will always exist, then the preimage of q consists of a finite collection of points p\in M. For each one we calculate the “signum” of f at p — written \mathrm{sgn}_pf — to be +1 if f_{*p} is orientation-preserving and -1 if f_{*p} is orientation-reversing. I say that

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

While we will not prove this until next time, there is one immediate consequence: if f is not onto, then any point in N\setminus f(M) has an empty preimage, and so we find that \deg f=0. In a sense, the degree of f is counting how many times the image of M covers that of N, considering orientation.

December 9, 2011 - Posted by | Differential Topology, Topology


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

    Comment by Rhonda ROO | December 9, 2011 | Reply

  2. […] 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 […]

    Pingback by Calculating the Degree of a Proper Map « The Unapologetic Mathematician | December 10, 2011 | Reply

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

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

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