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: 
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.)
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.