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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!

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    try to find a path from one solution to $F(z)=1$ to the next one along which the determinant of the derivative is nonzero. – 2017-01-29
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    @Thomas every path is such a path, because $z \in \mathbb S^1$, hence $z^{n-1} \neq 0$ but how does this help? – 2017-01-29

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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.