Map Surjective on a Disk I've got another question from a student that has stumped me:  Let $D^{n+1}$ be the $n+1$-disk, with boundary sphere $S^n$.  Suppose $f:D^{n+1}\longrightarrow \mathbb{R}^{n+1}$ is a map such that $f(S^n)\subseteq S^n$.  Furthermore, suppose that $f|_{S^n}$ has nonzero degree.  Show that $f(D^{n+1})$ contains $D^{n+1}$.  
I have to admit, I'm at a loss to even start this problem.
 A: Alright, under the general heading of the Hopf Degree Theorem, we have the Extension Theorem. I'm looking at Guillemin and Pollack, pages 145-146, in the smooth category actually.
BUT see WOOKIE for the continuous case: 
A map $f: \mathbb S^n \rightarrow \mathbb S^n$ is extendable to a map $F: \mathbb D^{n+1} \rightarrow \mathbb S^n$ if and only if $\deg(f)=0.$
The Extension Theorem is the same with the preimage replaced by any compact connected oriented $W$ with boundary $\partial W$ of dimension $n+1$ and $n.$   
Anyway, the  $f|_{S^n}$  you are given has nonzero degree, so there is no extension $F$ that maps all of the closed ball to the sphere. Meanwhile, assume that there is a point $U$ in the open ball that is not in the image of $f.$ Compose $f$ with central projection from $U$ onto $\mathbb S^n.$ This new map takes $\mathbb D^{n+1}$ to   $\mathbb S^n,$ so it is an extension. This is a contradiction.  
Note that it was not really necessary to have the missing point $U$ be at the origin, as the sphere is star-shaped around any point of the open ball. Furthermore, as the original $f\left(\mathbb D^{n+1}\right)$ is compact,  the distance from it to $U$ is bounded from below. This seems necessary for concluding that the composed map is continuous.  
