2

Let $F:M^n \to N^n$ be a proper smooth map between manifolds of dimension $n$. Then we get a map between compactly supported cohomologies $F^*: H^n_c(N) \to H^n_c(M)$ induced by pullback.

We define the degree of $F$ to be the constant $degF$ such that for any $\omega \in \Omega^n_c(N)$, $$\int_MF^*\omega=degF\int_N\omega$$

This is well-defined since the compactly supported cohomology of degree $n$ is isomorphic to $\Bbb{R}$ via integration, and so $F^*$ is given by multiplication by a constant.

I believe the following is true:

F is not surjective $\implies degF=0$

If we assume that $N-ImF=:V$ is an open set, then we can take any $\omega \in \Omega^n_c(N)$ such that $supp(\omega) \subset V$. Then $F^*\omega=0$, hence $degF=0$.

But I'm not sure how to prove the case when the unattained values in $N$ are a discrete set.

Tobyhas
  • 829

1 Answers1

2

The assumption that $F$ is proper guarantees that its image is closed, so that $N-\operatorname{Im} F$ is an open set. Exactly how you prove this depends on what your definition of "proper" is (for some definitions the condition that $F$ is a closed map is just part of the definition and for others you have to do some work and use the fact that $N$ is locally compact Hausdorff); see When is the image of a proper map closed? for instance.

Community
  • 1
Eric Wofsey
  • 342,377