does not has continuous extension of $S^1 \to S^1$ to $\Bbb{R}^2 \to S^1$ Let $S^1$ be the smooth manifold then the identity map is smooth $i:S^1 \to S^1$,prove there exist not continuous extension from $i:\Bbb{R}^2 \to S^1$
I try to construct some connected set in $\Bbb{R}^2$ to $S^1$ that fail to be connected,for example removing two point that still keeps the domain connected but not in $S^1$ is my idea correct?
 A: Suppose we may extend $i$ to $r$. That is $r:\mathbb{R}^2\rightarrow \mathbb{S}^1$ is such that $r|_{\mathbb{S}^1}=i$ and is continuous. So in particular, $r$ is a retraction from $\mathbb{R}^2$ to $\mathbb{S}^1$. Thus $\mathbb{S}^1$ is a retract of a simply connected space, which implies that $\mathbb{S}^1$ is simply connected. Contradiction.
A: Suppose such extension $f:\mathbb{R}^2\rightarrow S^1$, exist, $i\circ f$ is the identity, implies that $(i\circ f)_*:\pi_1(S^1)\rightarrow\pi_1(S^1)$ is the identity., we have $(i\circ f)_*=i_*\circ f_*$ and $f_*$ is constant, contradiction since $\pi_1(S^1)=\mathbb{Z}$.
A: Suppose $i:\mathbb{S}^1\rightarrow \mathbb{S}^1$ is continuous map admitting a continuous extension $\tilde{i}:\mathbb{R}^2\rightarrow \mathbb{S}^1$ . Then, in particular, $\tilde{i}|_{\mathbb{S}^1}=i$. Hence, $\mathbb{S}^1$ is a retract of $\mathbb{R}^2$. However, this implies that
$0=H_1(\mathbb{R}^2)=H_1(\mathbb{R}^2,\mathbb{S}^1)\oplus H_1(\mathbb{S}^1)=H_0(\mathbb{S}^1)\oplus H_1(\mathbb{S}^1)=\mathbb{Z}^2$
