Another supposedly easy question concerning the degree of mapping.
Let $M$, $N$, be orientable manifolds of equal dimension, $M$ compact, $N$ connected and $F : M \rightarrow N$ a mapping. The degree of $F$, is defined as $$\deg(F) := \sum_{p \in F^{-1}(\{q\})} \operatorname{sgn}( \det (d F_p)),$$ for a regular value $q \in N$ of $F$.
(as for the degree mod 2, this is welldefined since the degree is invariant under homotopy and of the chosen regular value)
Consider the map $F: \mathbb S^1 \rightarrow \mathbb S^1 \subset \mathbb C$, $z \mapsto z^n$.
My textbook says $deg(F) = n$.
How do I get this result?
I looked at it like this: $d F_z = n \, z^{n-1}$, and pick $q=z^n=1$ wlog.
but then what is sgn($z^{n-1}$) can be positive and negative (according to the complex sgn function), or not?
any help is appreciated!
Please correct me if I'm wrong, I think it works like this:
Pick $z=1$ as your first regular point.
sgn(d$F_1$) = sgn$(n) =+1, \quad (n \in \mathbb N) $
Now, as pointed out, the determinant along any path is nonzero. The determinant is a continuous function, hence the sign doesn't change.
Therefore, it will be positive for all regular points.