Are $f(x)=2x+1$ and $g(x)=\frac{2x-1}3$ topologically conjugate on $\Bbb Z_2$?

Question

Let $f,g$ be functions from $\Bbb Z_2\to\Bbb Z_2^\times$

$f(x)=2x+1$ and $g(x)=\frac{2x-1}3$

Then I have the notion that $f,g$ are topologically conjugate to each other in the 2-adic metric. I'd like to verify that but I find it offputting that the domains of the functions don't coincide with the range.

Anyway, proceding regardless, I can ask myself "is there a homeomorphism that conjugates one to the other?"

So I guess the homeomorphism would be $x\mapsto\frac{x-2}3$ or $x\mapsto3x+2$

So to my tiny brain, that homeomorphism looks good because it's one to one and onto (because the maps to and from zero fall outside of $\Bbb Z_2^\times$, preserves the 2-adic value $\lvert x\rvert_2=1$ and $d(f(x),f(y))=d(x,y)$. Is that correct?

I'm concerned I don't have any conjugation going on there, just a transformation from one to the other. But how could I, as the codomain is not within the range?

EDIT:

I've moved my thinking along. If I let $h=\frac{x-2}3$ and conjugate $f$ I seem to get $hfh^{-1}=f$ rather than $g$ so I'm definitely doing something wrong here. Note, I have no doubt they are topological conjugates, but I want to derive the homeomorphism.

The fact $hfh^{-1}=f$ suggests $f,h$ commute - so i checked whether $ax+b$ and $cx+d$ commute in general, and I got $hfh^{-1}=ax+(1-a)d+bc$, confirming they DON'T in general commute, but I substituted in this special case and got $ax+(1-a)d+bc=2x+1$. I found this which looks relevant but I seem to be going down a rabbit hole not relevant to the original question.

Answer 1

It turns out that $f$ and $g$ actually are topologically conjugate.

Let's simplify the question first a bit by conjugating the two functions with the translation $t(x)=x+1$. We get $$ \tilde{f}(x):=t(f(t^{-1}(x)))=(2(x-1)+1)+1=2x, $$ and $$ \tilde{g}(x):=t(g(t^{-1}(x)))=([2(x-1)-1]/3)+1=2x/3. $$ Topological conjugacy is an equivalence relation, so $f$ and $g$ are conjugate if and only if $\tilde{f}$ and $\tilde{g}$ are.

Let us then define the function $h:\Bbb{Z}_2\to\Bbb{Z}_2$ by the recipe $$ h(x)=\begin{cases}0,&\text{if $x=0$, and}\\ 3^{\nu_2(x)}x,&\text{otherwise.}\end{cases} $$ Here, for all $x\neq0$, $\nu_2(x)=\ell$, when $x\in 2^\ell\Bbb{Z}_2\setminus2^{\ell+1}\Bbb{Z}_2$. Then

• As $3$ is a $2$-adic unit, and $\Bbb{Z}_2$ is the disjoint union of the subsets $U_\infty:=\{0\}$ and $U_\ell:=2^{\ell}\Bbb{Z}_2\setminus 2^{\ell+1} \Bbb{Z}_2$, $\ell=0,1,2,\ldots$, it follows that $h$ maps each subset $U_\ell, \ell=0,1,\ldots,\infty$, bijectively onto itself. Therefore $h$ is itself also a bijection.
• If $V=x+2^\ell\Bbb{Z}_2$ is any basic open subset of $\Bbb{Z}_2$, then $$h^{-1}(V)=h^{-1}(x)+2^\ell\Bbb{Z}_2$$ is another basic open subset. Therefore $h$ is continuous.
• The argument of the previous bullet holds equally for the obvious inverse of $h$ (replace $3$ by $1/3$ everywhere), so we can conclude that $h$ is a homeomorphism.
• Finally, we also have, for all $x\neq0$ in $\Bbb{Z}_2$ $$h(\tilde{g}(x))=3^{\nu_2(2x/3)}(2x/3)=3^{1+\nu_2(x)}(2x/3)=3^{\nu_2(x)}\cdot 2x=\tilde{f}(h(x)).$$ So $h$ gives the topological conjugacy between $\tilde{f}$ and $\tilde{g}$ proving the main claim also.