$z \mapsto z^m$ is orientation preserving and regular

60 Views Asked by At

I'm following GP's book of Differential Topology and I stuck in the proof of Fundamental Theorem of Algebra using Intersection Theory.

We have the afirmation of $f: \mathbb{S^1} \to \mathbb{S}^1, s.t., z \mapsto z^m$ (m > 0) be a regular and orientation preserving map. Then everything follows from the proposition:

"Let $f: X \to Y$ smooth, $X = \partial W$, for W a compact manifold, Y a connected manifold, with dim(X) = dim(Y). So, if $deg(f) \neq 0$, hence f does not admit a extension $:W \to Y$".

My problem was only to show that f is regular and orientation preserving, I understand how the FTA follows from the proposition.

For f regular my attempt is:

Given $z \in \mathbb{S}^1$, the differential in z, $D := df_z: T_z \mathbb{S}^1 \to T_{z^m} \mathbb{S}^1$, is a linear map, not trivial.

Then from Kernel-Image:

$1 = dim(T_z \mathbb{S}^1) = dim (Im(D)) + dim (Ker(D))$, with $dim (Im(D)) > 0$.

So $dim (Im(D)) = 1 = dim(T_{z^m} \mathbb{S}^1)$.

Then $D$ is surjective (indeed, will be a linear isomorphism).

But I have no ideia in how to prove that this maps preserves orientation (maybe beacause I do not have sufficient maturity in this topic).

For f preserves orientation the definition is: Given $z \in \mathbb{S}^1$, the differential $df_z$ has positive determinant (equivalentely, $df_z$ carrys positive ordenated basis to positive ordenated basis).

The book give a clue of use the local parametrization $\phi: \theta \mapsto (\cos \theta, \sin \theta)$, but i don't know how to apply this.