How can I find a lift for this circle homeomorphism?

Question

Let me consider the circle homeomorphism $T:\Bbb{R}/\Bbb{Z}\rightarrow \Bbb{R}/\Bbb{Z}$s.t. $T(x)=x+\frac{1}{2}$ if $x\in [0, 1/2]$ and $T(x)=x-\frac{1}{2} $ if $x\in [1/2,1]$.

Now I want to find a lift $F:\Bbb{R}\rightarrow \Bbb{R}$. This means $F$ needs to be continuous and $\pi(F(x))=T(\pi(x))$ for all $x\in \Bbb{R}$ where $\pi:\Bbb{R}\rightarrow \Bbb{R}/\Bbb{Z}; x\mapsto x+\Bbb {Z}$.

My idea was somehow to glue this two functions together. But the problem is that I don't see how to guarantee that $\pi(F(x))=T(\pi(x))$ holds. I thought about defining $F(x)=x+\frac{1}{2}$ if $0\leq x~ \operatorname{mod}~ 1< 1/2$ and $F(x)=x- \frac{1}{2} $ if $1/2\leq ~ \operatorname{mod}~ 1 \leq 1$ . Equivalently I can say

$$F (x) = \left\{ \begin{array}{ll} x+\frac{1}{2} & x \in [k, k+1/2), \forall k\in \Bbb{Z} \\ x- \frac{1}{2}& \, x\in [k+1/2, k+1], \forall k\in \Bbb{Z} \\ \end{array} \right. $$

But I think I messed up with continuity of $F$.

Can someone tell me if it works and if not correct my idea?

Thanks a lot

Answer 1

Simply take $F(x) = x + \frac 1 2$. Then we have $F(x) - F(y) = x - y$ for all $x, y \in \mathbb R$. Writing $\pi(x) = [x] = x + \mathbb Z$, we therefore see that if $[x] = [y]$, then $[F(x)] = [F(y)]$. This means that $F$ induces a unique $F^* : \mathbb R/ \mathbb Z \to \mathbb R/ \mathbb Z$ such that $\pi \circ F = F^* \circ \pi$. Explicitly, $F^*([x]) = [F(x)] = [x + \frac 1 2]$.

It remains to show that $F^* = T$. Since each $\xi \in \mathbb R/ \mathbb Z$ can be written as $\xi = [x]$ with $x \in [0,1]$, it therefore suffices to consider the following two cases:

1. $x \in [0, \frac 1 2]$. Then $$F^*([x]) = [x + \frac 1 2] = T([x]).$$

2. $x \in [\frac 1 2, 1]$. Then $$F^*([x]) = [x + \frac 1 2] = [x + \frac 1 2 - 1 ] = [x - \frac 1 2] = T([x]).$$

Answer 2

Hint: in $\Bbb{R}/\Bbb{Z}$, you are working modulo $1$, just as in $\Bbb{Z}_2$ (a.k.a., $\Bbb{Z}/\Bbb{2Z}$) you work modulo $2$. In modulo $2$ arithmetic, you have $-1 \simeq 1$, because $1 - (-1) = 2 \simeq 0$, and similarly, in modulo $1$ arithmetic you have $-\frac{1}2 \simeq \frac{1}{2}$. So you can forget all those minus signs: your function $T$ is just adding $\frac{1}{2}$ modulo $1$.