Show that there exist a retraction ($r:D^n\rightarrow S^{n-1}$, such that the restriction is the identity in $S^{n-1}$) if and only if $S^{n-1}$ is contractible.
I know from another exercise that if $S^{n-1}$ is contractible then there exist a Homotopy $F:S^{n-1}\times I\rightarrow S^{n-1}$ between $id_{S^{n-1}}$ and a constant map lets say $c:S^{n-1}\rightarrow S^{n-1}$, meaning
$$F(x,0)=id_{S^{n-1}}(x)=x$$ $$F(x,1)=c(x)=x_0$$
I also know that there exist a natural continuous function $g:F(x,0)=S^{n-1}\times I\rightarrow D^n$ defined by $g(x,t)=xt$.
How can I define $r$? Also, I have no idea on how to prove the other direction. thanks in advance.
Let $F$ be the homotopy $F:S^{n-1}\times I\to S^{n-1}$ from the identity to the constant map. Note then that we may factor $F$ through the map $g:S^{n-1}\times I\to D^n$ which is the other such map you mentioned. Another way to think of $g$ is a quotient map, quotienting $S^{n-1}\times \{0\}$, which gives the fact that you can factor $F$ as a consequence of it being constant on that subspace.
We therefore have the commutative diagram: $\require{AMScd}$ $$ \begin{CD} S^{n-1}\times I @>F>> S^{n-1}\\ @VgVV @|\\ D^n @>r>> S^{n-1} \end{CD}$$
The map $r$ is the one you are looking for.
For the other direction, take$F$ to be $r\circ H\circ (i\times id_I)$, where $i$ is the inclusion of $S^{n-1}$ into $D^n$, and $H$ is any contraction of $D^n$.