# 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.

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Maybe assume that there is a point $a \in f(D^{n + 1}) \setminus D^{n + 1}$, then $a - f(x) \neq 0$, .... – Dylan Moreland Jul 17 '12 at 23:11
Actually, I don't see where that's going either :-/ – Dylan Moreland Jul 17 '12 at 23:19
Assume that the interior point missed is the origin, if necessary by an LFT. Then central projection makes a continuous map onto $\mathbb S^n.$ If not surjective, it is homotopic to a constant map. Well, it's a start. – Will Jagy Jul 17 '12 at 23:48
It has to be surjective, because the map has nonzero degree; and that's the contradiction. – user641 Jul 17 '12 at 23:50
@Will: Perhaps you (or SteveD) should post this as an answer. – Jason DeVito Jul 18 '12 at 1:40

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.

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