Is the image of a contractible space under a contraction contractible?

Question

Is the image of a contractible space under a contraction contractible?

More specific to what I'm looking for: is the image of the unit ball $B_1(0)\subset\mathbb{R}^n$ under a contraction mapping $F:\mathbb{R}^n\to\mathbb{R}^n$ still contractible?

Answer 1

No. For instance, the map \begin{align*} F: \mathbb{R}^2 &\to \mathbb{R}^2 \\ (x,y) &\mapsto \frac{1}{4\pi}\left(\cos(2\pi x), \sin(2\pi x)\right) \end{align*} takes the unit ball to a circle, which is not contractible. That the map is a contraction can be seen by the derivative, or by noticing that it is the composition of the projection in the first coordinate with a walk on the circle with constant velocity equal to $1/2$. (This should make for a quick visualization that it is indeed a contraction.)

Answer 2

The map $H_t(x)=tF(x), t\in [0,1]$ restricted to the unit ball defines a deformation retract of the unit ball to a point.

Answer 3

Consider the smooth surjective exponential map, which is Lipschitz continuous $e: T_S S^n = \mathbb{R}^n \rightarrow S^n$ ($S$ being the south pole).

Consider a large $M>0$ and a small $r > 0$ and define $F(x_1, \ldots, x_{n+1})=r \cdot e(Mx_1, \ldots, Mx_n) \in S^n \subset \mathbb{R}^{n+1}$.

If $M$ is large enough, then $F(B_1(0))=rS^n$ for any $r$. If you choose then $r$ small enough, $F$ is a contraction, and the image of $B_1(0)$ is homeomorphic to $S^n$, which is not contractible.