Proposition: There is no continuous map from the unit disc $D^2$ to its boundary $S^1$ whose restriction to $S^1$ is the identity on $S^1$.
My proof: Assume that there is such an $f$. Let $g: S^1\to S^1$ be a continuous map. Then, $g\circ f: D^2 \to S^1$ is an extension of $g$ to $D^2$. Hence $\pi_1(S^1)=0$. Contradiction.
The simplicity of my solution makes me suspicious. Is everything correct?
In fact, there is no retraction $r: D^2 \rightarrow S^1$. For if there was, the induced homomorphism of fundamental groups of the inclusion map $j: S^1 \rightarrow D^2$ would be injective. But since the fundamental group of $D^2$ is trivial, while the fundamental group of $S^1$ is not, this contradicts the induced homomorphism of the inclusion map being injective. Therefore, no such retraction $r$ exists.