Prove that a square without one inner point is not simply connected

Question

I understand that this is a consequence of Brouwer's theorem. For a circle without an inner point, we have that the circle's mapping to itself does not contract, i.e. if $f = \mathrm{id}\colon S^1 \to S^1$, $f(x) = x$, then $F\colon D^2 \to S^1$ extending $f$ does not exist. But how to prove it for a square is not clear.

Answer 1

Let $S = \{(x, y) \in \mathbb{R}^2 \mid -1 \leq x, y \leq 1\}$ be the closed square and $D = \{(x, y) \in \mathbb{R}^2 \mid \sqrt{x^2 + y^2} \leq 1\}$ the disk. A homeomorphism $f\colon S \to D$ is given by $f(0) = 0$ and $$ f((x, y)) = \frac{\max\{|x|, |y|\}}{\sqrt{x^2 + y^2}}(x, y) $$ for all $(x, y) \neq (0, 0)$ (see this answer for a justification), and since $f$ takes interior points to interior points as well as boundary points to boundary points, your problem reduces to showing that $D$ minus an interior point is not simply connected.