Non-surjective map $F$ implies $\int_M F^* \eta=0$

Question

This is a problem from Lee 17.12:

Suppose $M$ and $N$ are compact, oriented, smooth n-manifolds, and $F:M\rightarrow N$ is a smooth map. Prove that if $\int_M F^*\eta \neq 0$ for some $\eta \in \Omega^n(N)$, then $F$ is surjective.

I want to show the contrapositive, that non-surjective map implies $\int_M F^*\eta = 0$, and we know $\int_M F^*\eta = \text{deg}(F)\int_N\eta$. But I am not sure why non-surjectivity implies that $\text{deg}(F)=0$. Please help!

PS: I prefer to use techniques from Lee, as this is a problem there.

Answer 1

OK so I think this one is actually easy: Let $p \notin \text{Im}(F)$, then note that $p$ is a regular value (I didn't actually notice this before), so use this formula of degree instead: $\text{deg}(F)=\sum_{x\in F^{-1}(q)} \text{sgn}(x)$, but $F^{-1}(q)$ is an empty set so $\text{deg}(F)=0$. Thus non-surjective map gives zero degree.