I'd like to know if my argument below, in which I try to show that the 1-sphere is not a retract of the 2-disk using homotopy, is valid.
Suppose there is a retract $r:D^2 \to S^1$. Then we can consider the induced homomorphism $r_\ast:\pi_1(D^2,x_0) \to \pi_1(S^1,x_0)$, where the basepoint $x_0 \in S^1 \subset D^2$. Let $[f] \in \pi_1(D^2,x_0)$, where $f:I \to D^2$ maps $I$ to $S^1 \subset D^2$. Now we have $r_\ast[f]=[rf] \in \pi_1(S^1,x_0)$. Since $r$ fixes $S^1$, it follows that $r_\ast[f]$ is nontrivial. But $\pi_1(D^2,x_0)=0$ and $\pi_1(S^1,x_0)\cong \mathbb{Z}$, so there cannot be a retract $r:D^2 \to S^1$.
Does this work? Thanks.
Usually, we can get past needing to use explicit elements of the fundamental group by using the 'composition' definition of a retract. The map $r\colon X\to A$ is a retract for the subspace $A\subset X$ if $r\circ i=\mbox{Id}_A$ where $i\colon A\to X$ is the inclusion map.
Now, suppose there is a retract $r\colon D^2\to S^1$, then $r\circ i=\mbox{Id}_{S^1}$ and so $r_*\circ i_*={\mbox{Id}_{S^1}}_*$ by the functorality of $\pi_1$. Also by functorality, we know that ${\mbox{Id}_{S^1}}_*=\mbox{Id}_{\pi_1(S^1)}$ but note that we have $r_*\colon 1\to \mathbb{Z}$ which must be the trivial map and so $r_*\circ i_*$ must also be the trivial map. The identity map on $\pi_1(S^1)$ is not the trivial map because $\pi_1(S^1)$ is not trivial - contradiction.