I'm looking for a proof of a statement similar to the following: let $\phi : S^2 \to S^2$ be a smooth function with $\mbox{deg}\,\phi \neq 0$. Then any extension $\Phi$ of $\phi$ to the closed ball $\overline{B^3} \subset \mathbb R^3$ has at least an interior discontinuity.
To be more precise, my trouble with this statement is the fact that the discontinuity must lie in the open ball $B^3$. In particular, I was wondering why $\Phi$ cannot be discontinuous, as a function on $\overline{B^3}$, on $S^2$ (althought, restricted to $S^2$, it is smooth). Note that, when $\phi = id$, as a corollary of Lemma 5 in §2 of Milnor's Topology from the differentiable viewpoint, the quoted author says that the identity map $id : S^2 \to S^2$ cannot be extended to a smooth map $\overline{B^3} \to S^2$, no mention of the interior.