specific formula of a continuous map

Question

Let $X= \mathbb{S}^1\cup \{(x, 0): -1<x<1\}$, where $\mathbb{S}^{1}=\{(x, y)\in \mathbb{R}^2: x^2+y^2=1\}$. Let $f:X\to X$ be a continuous map such that $f((-1, 0))= (-1, 0)$ and $f((1, 0))= (1, 0)$. Also, for any $(x, y)\in \mathbb{S}^1\backslash \{(-1, 0),(1, 0)\}$, the first coordinate of $f((x, y))$ is larger than $x$. Also, for any $(x, y)\in \{(x, 0): -1<x<1\}$, the first coordinate of $f((x, y))$ is less than $x$. In my research I need to give the specific formula of $f$. Please help me to know it.

Answer 1

There are many ways to construct such a function; here’s one straightforward one to get this off the unanswered list.

Let $s:[-1,1]\to\Bbb R:x\mapsto\frac12(1-|x|)$; then $-1\le x-s(x)\le x+s(x)\le 1$ for any $x\in[-1,1]$, and $s(-1)=s(1)=0$. It’s easily checked that the function

$$f:X\to X:\langle x,y\rangle\mapsto\begin{cases} \langle x-s(x),0\rangle,&\text{if }y=0\\ \left\langle x+s(x),\sqrt{1-\big(x+s(x)\big)^2}\right\rangle,&\text{otherwise} \end{cases}$$

has the desired properties.