# Show that if $t : \Bbb T →\Bbb T$ is continuous, then we may write $t(θ) = pθ + P (θ)$, where $P : \Bbb R →\Bbb R$ is a $\Bbb Z$-periodic function

A function $P : \Bbb R → \Bbb R$ is $\Bbb Z$-periodic if $P (x + n) = P (x)$, for all $x ∈ \Bbb R, n ∈ \Bbb Z$. Suppose that $T : \Bbb R → \Bbb R$ is continuous and satisfies $T(x + n) − T(x) ∈ \Bbb Z$, for all $x ∈\Bbb R, n ∈\Bbb Z$. Show that there exists $p ∈\Bbb Z$ and a continuous $\Bbb Z$-periodic function $P :\Bbb R →\Bbb R$ such that $T (x) = px + P (x)$, for all $x ∈\Bbb R$.

Show that if $t : \Bbb T →\Bbb T$ is continuous (where $\Bbb T = [0, 1]$ with $0$ and $1$ identified), then we may write $t(θ) = pθ + P (θ)$, where $P : \Bbb R →\Bbb R$ is a continuous $\Bbb Z$-periodic function, and $p ∈\Bbb Z$. (For this part, you may assume that every continuous map $t:T→T$ lifts to a continuous map $T :\Bbb R→\Bbb R$.)

Show that a necessary condition for $t$ to be a homeomorphism is $p = ±1$. Is this condition sufficient? Why/Why not? (Hint: Consider $|T (1) − T (0)|$.)

Answer: Now for the first part considering $f_n(x)=T(x+n)-T(x): \Bbb R \to \Bbb Z$ continuous we have that $f_n(x)=d_n \in\Bbb Z$. Then we can derive by induction that $d_n=nd_1$. Now considering $d_1=p$ we will get the final expression that $T (x) = px + P (x)$.

For the next part if I want to extend $t$ from $\Bbb T$ to $\Bbb R$ then we see that $t(0)=t(1)$(mod $1$) i.e $t(0)-t(1)\in \Bbb Z$ (say, $p$) then considering $P(\theta)=t( \theta)- p\theta$. How can I extend it to $\Bbb R$?

For the third part if $p \neq \pm 1$ then I am getting that $t(\theta)=p\theta+P(\theta)$ and $|p|>1$ say then $t$ is homeomorphism in $\Bbb T$ then $exp(2\pi i p \theta + P(\theta)$ will be homeomorphism in $S^1$. So, I was putting $\theta=0, \frac 1p$ but do not get anything. Please help.

Let $[x]$ be the integer part of $x$. For example, for an orientation-preserving homeomorphism you can define the continuous map $$T(x)=t(x-[x])+[x].$$
For the third part, it suffices to consider $x=1$ (which is the same as $x=0$). Note that the equation $t(x)=1$ has $|p|$ solutions and so it cannot be invertible when $|p|>1$. The other direction is direct.
The number $|p|$ of solutions follows from solving $p\theta=0 \bmod 1$, which gives $\theta=k/p$ with $k=1,\ldots,p$.
• For the second part. Is the map $T$ continuous? Jan 8, 2018 at 12:03
• And for the third part why it would have $|p|$ solutions? Could you please write it down explicitly? Jan 8, 2018 at 13:10
• But I have $t(\theta)=p\theta + P(\theta)$ so what about $P(\theta)$ part? In my question also I have mentioned that I am stuck with that part. Jan 9, 2018 at 11:32
• But I have $t(\theta)=p\theta + P(\theta)$ so what about $P(\theta)$ part? In my question also I have mentioned that I am stuck with that part. For $t(\theta)=1 \Rightarrow p\theta +P(\theta)=1$. Now putting $\theta=k/p$ I have $k+P(k/p)=P(k/p)$(mod $1$). Now? Jan 9, 2018 at 11:38