Dtyjlui

Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$

I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$

I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.

Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).

Let us proceed in various steps.

(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.

(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.

(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.

(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$