3
$\begingroup$

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?

  • 3
    What is your definition of degree? – 2017-01-28
  • 1
    good point! i updated the question. – 2017-01-28
  • 4
    The degree mod $2$ is something that does not depend on the regular value at which it is evaluated (provided the space $N$ is connected obviously). Since points not in the image are regular values and have degree $0$ the degree of the entire map is $0$. – 2017-01-28
  • 0
    Do you know the proof that the degree is $0$ for the special case that $M$ and $N$ are spheres (of the same dimension)? If I remember correctly, the argument is easier to follow along in that case. – 2017-01-28
  • 1
    _"Since points not in the image are regular values..."_, thanks @s.harp! i completely forgot about – 2017-01-28

1 Answers 1

3

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 $\deg_2(F) = |\emptyset| \mod 2 = 0$.