Finding a retract on compact subset of $\mathbb{R}^2$

Question

Is it possible to find a retract $r:K \to M$ where the domain $K \subseteq \mathbb{R}^2$ is compact and convex and the codomain is given by $M = [-1,1] \times \{0\} \ \cup \{0\} \times [-1,1] \subseteq K$

Answer 1

Define

$$A=\{(x,y)\ |\ x\geq y\geq 0\}$$ $$A_0=\mathbb{R}\times 0$$

Lemma 1. There's a retraction $f:A\to A_0$ such that $f(x,x)=(0,0)$ for any $x$.

Proof. Let $g_\alpha:\mathbb{R}^2\to\mathbb{R}^2$ be the rotation by angle $\alpha$. Let $p:\mathbb{R}^2\to\mathbb{R}^2$ be the projection onto X-axis. Now for a given $(x,y)\in A$ take it's angle with X-axis, namely $\beta(x,y)=\arcsin(\frac{y}{\lVert(x,y)\rVert})$. By the definition $0\leq \beta(x,y)\leq \pi/4$ so now define

$$f(x,y)=p(g_{\beta(x,y)}(x,y))$$

I leave it as an exercise that $f$ is the desired function. $\Box$

Lemma 2. There's a retraction $F:\mathbb{R}^2\to \mathbb{R}\times 0\cup 0\times\mathbb{R}$.

Proof. Take a retraction from Lemma 1. Now you perform the same operation $8$ times: rotate $f$ by $\pi/4$ and take mirror of it. Both properties of $f$ ensure that the resulting function agrees on itersections and thus continuous. It is the desired retraction. I leave details as an exercise. $\Box$

Corollary 1. $[-1,1]\times 0\cup 0\times [-1,1]$ is a retract of $\mathbb{R}\times 0\cup 0\times \mathbb{R}$ and therefore is a retract of $\mathbb{R}^2$.

Proof. The first part is easy and the second part even simplier. :) Thus I leave it as an exercise. $\Box$

Corollary 2. $[-1,1]\times 0\cup 0\times [-1,1]$ is a retract of any subset of $\mathbb{R}^2$ containing it.

Proof. Obviously if $f:\mathbb{R}^2\to[-1,1]\times 0\cup 0\times [-1,1]$ is a retraction then all you have to do is restrict the domain. $\Box$

Answer 2

Let $K$ be the convex hull of $M$. Then clearly $K$ retracts (actually, deformation retracts) onto $M$. Just push the sides towards the origin.

In fact any convex set $K'$ which contains $M$ deformation retracts onto $M$. Sketch: $K'$ is homeomorphic to a disk so it deformation retracts onto its sub-disk $K$.