Equivalence homotopy between $\Bbb R^2$\-{$(0,0)$} and a convex set without a point

Question

I can't to show the equivalence homotopy between $\Bbb R^2$-{$(0,0)$} and a convex set( in $\Bbb R^2$) without the point $(0,0)$ where $(0,0)$ belongs to the interior part of the convex set. I have to find two functions $i$ and $f$ such that $f \circ i$ is homotopy to the identity of convex set without $(0,0)$ while $i \circ f $ is homotopy to the identity of $\Bbb R^2$-{$(0,0)$}. The map $i$ can be the immersion of the convex set in $\Bbb R^2$-{$(0,0)$} but $f$?

Answer 1

Ok, so let $A\subseteq\mathbb{R}^n$ be a convex set and $v\in int(A)$ be a fixed vector. Without a loss of generality we may assume that $v = 0$ (because if it is not then we can just "move" the entire set via the homeomorphism $x\mapsto x-v$). So now since $0\in int(A)$ then there exists an open ball around $0$ contained in $int(A)$. And if there exists an open ball, then there also exists a closed ball $D$ (being a subset of the open ball) contained in $A$ centered in $0$ with radius $r$.

Now define

$$H:A\times I\to A$$ $$H(a, i) = \begin{cases} a & \mbox{if } a\in D\\ (1-i)\cdot a + i\cdot r\cdot\frac{a}{\lVert a\rVert} & \mbox{otherwise } \end{cases}$$

Note that the map is well defined (i.e. the right side is always in $A$) because $A$ is convex. It is also continous because values (of upper and lower definition) agree on the common set of arguments (i.e. the boundary of $D$). So what I'm doing here is I'm taking any point outside the ball $D$ and I'm deforming it onto the boundary of $D$ while leaving all points of $D$ in place.

This mapping is a deformation retract between $A$ and the closed ball $D$. In particular $A$ is homotopy equivalent to $D$. Now this homotopy can be refined. Note that for any $i$ we have $H(0, i)=0$. In particular you can remove $0$ from both domain and codomain to obtain a homotopy equivalence between $A-0$ and $D- 0$.

Now $A$ was picked arbitrarly so if I take $A=\mathbb{R}^n$ I obtain that $\mathbb{R}^n-0$ is homotopy equivalent to some closed ball minus center as well. So all that is left to prove is that two closed balls withou the center are homotopy equivalent. But simple translation and scaling gives not only homotopy equivalence but also a homeomorphism.