Degree of a restriction of a continuous map?

Question

I have a map $f:D^2 \rightarrow S^2$ and $f(-x)=-f(x)$ for $x \in S^1$. Does this mean that $\deg(f|_{S^1})=0$? if so, why? We defined this degree on $S^1$ as $f(\exp(t))=\exp(F(t))$ then $\deg(f)=F(1)-F(0)$. I also know that $\pi_1(D^2)=0$ and $\pi_1(S^1) =\mathbb{Z}$, if this is of any help.

Answer 1

Every map from $D^2$ to $S^1$ is null-homotopic because $D^2$ deformation retracts onto a point, and so in particular so is the restriction of such a map to a subspace of $D^2$. A null-homotopic map from the circle to itself has zero-degree as the constant map has zero degree and degree is a homotopy invariant.

It is in fact also the case that a map $f\colon S^1\to X$ is null-homotopic if and only if $f$ can be extended to a map $\tilde{f}\colon D^2\to X$ such that $\tilde{f}|_{S^1}=f$.