not surjective implies degree 0 This seems like an easy question, but I can't get my head around it.

Consider manifolds $M$, $N$, $F \in C^\infty(M,N)$, $\dim(M) = \dim(N)$, $M$ compact, $N$ connected. 
  The degree of $F$ is defined as the number of points in the preimage of some regular value $q$ of $F$ in $N$, that is $$\deg_2(F) := \operatorname{card}(F^{-1}(\{q\})) \mod 2.$$

(This is well defined by homotopy invariance.)
Why is $\deg_2(F)=0$ if $F$ is not surjective?

Consider $N \setminus F(M)$. Since $F$ is not surjective, this set is non-empty. By definition, $q \in N \setminus F(M)$ is a regular value (it has no pre-image) and $F^{-1}(q) = \emptyset$ so $\deg_2(F) = |\emptyset| \mod 2 = 0$.
