$G$ must hit the origin. - Is this proof legit?

Question

The map $F : \mathbb{R}^3 \to \mathbb{R}^3$ given by $F(x, y, z)=(-z, -x, -y)$ restricts to a map $f : S^2 \to S^2$ from the 2-sphere to itself.

Show that if $G$ is another map of Euclidean space to itself that restricts to $f$ on the 2-sphere, then $G$ must hit the origin.

Assume for contradiction that $G$ does not hit the origin, then $G/|G|$ will extend to $F$ in the whole disk. Implying that $G/|G|$ has degree zero according to the proposition. Hence $F$ has degree 0, contradicts with the fact that $\deg F = -1.$

Proposition. Suppose that $f: X \to Y$ is a smooth map of compact oriented manifolds having the same dimension and that $X = \partial W$ ($W$ compact). If $f$ can be extended to all of $W$, then $\deg(f) = 0$.

Thank you~~~~~

Answer 1

I think you are thinking right, but I would change some words:

Assume for contradiction that $G$ does not hit the origin, then $G/|G|$ is a function from the whole disk to the sphere. Implying that $f=(G/|G|)_{|S^2}$ has degree zero according to the proposition, contradicting with the fact that $\deg f = -1.$

Also note that in this case you don't need the general proposition: it's just the observation that a function from a sphere is contractible iff it extends to the disk. And if contractible then it is of degree $0$.