Almost retracting a ball to its boundary

Question

One cannot continuously retract a closed ball $B_n$ to it boundary $S_n$, that is one cannot find a continuous map $f:B_n\to S_n$ such that $f|_{S_n} = \mathrm{id}_{S_n}$. What if we drop the latter condition and allow $f$ to move points on $S_n$ yet be onto?

Answer 1

How about the ball in $\mathbb{R}^2$?

Let $f(re^{i\theta})=e^{2ri\theta}$ for $\theta\in[0,\pi]$ and $f(re^{i\theta})=e^{(4\pi-2\theta)ri}$ for $\theta\in[\pi,2\pi]$, where $0\leq r\leq 1$.

EDIT: The intuition is that, on the circle $S_1$, the map is what you get by adding a path going round the circle once and its inverse in the fundamental group. Then use the shrinking radius $r$ to shrink the path down to being constant.

In fact, by using $\pi_n(S_n)\neq0$, you can apply this method to $B_n$ for any $n$.