Proposition

Prove that the circle is not a retract of the closed disk.

Solution

Let $\overline{D} = \{ z \in \mathbb{C} \mid \abs{z} \leq 1 \}$ and $S^1 = \{ z \in \mathbb{C} \mid \abs{z} = 1 \}$.

We will first show that $\overline{D}$ is simply connected.

$\overline{D}$ is path-connected. We will check if $\pi_1(\overline{D}, 0)$ is trivial. Let $[f] \in \pi_1(\overline{D}, 0)$ be given. Consider $H: I \times I \rightarrow \overline{D}$ such that $H(s, t) = (1 - t)f(s)$.

• $H$ is continuous.
• $\forall s \in I, H(s, 0) = f(s)$.
• $\forall s \in I, H(s, 1) = 0$.
• $\forall t \in I, H(0, t) = (1 - t)f(0) = 0$.
• $\forall t \in I, H(1, t) = (1 - t)f(1) = 0$.

Therefore, $f$ is path-homotopic to $e_0$ where $e_0$ is the constant function at $0$. Thus $\pi_1(\overline{D}, 0) = \{ [e_0] \}$, so $\overline{D}$ is simply connected.

A retract of a simply connected space is simply connected. (See Corollary 7.29 on P.198 of Introduction to Topological Manifolds) Since $\overline{D}$ is simply connected and $S^1$ is not, $S^1$ cannot be a retract of $\overline{D}$.