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

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~~~~~

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 $0$.