# Is $(2x-1)/3$ a homeomorphism on $\Bbb Z_2$? And does it expand or contract?

Question

Is $$f(x)=(2x-1)/3$$ a homeomorphism on the $$2$$-adic integers $$\Bbb Z_2$$? And does it expand or contract?

My attempt

$$\lvert(2x-1)/3-(2y-1)/3\rvert_2=\lvert(2x-2y)\rvert_2=\frac12\lvert x-y\rvert_2$$ so I'm fine on it preserving the topology, it just shrinks stuff.

In particular, I'm asking about the fact it doesn't surject. It maps from $$\Bbb Z_2\to\Bbb Z_2^\times$$

Also, I'm confused because it shrinks the pair $$x,y$$ but $$\Bbb Z_2\to\Bbb Z_2^\times$$ looks like it's growing stuff because it would take a ball of radius $$1/2$$ around zero to a ball of radius one.

I'm probably making a schoolboy error, please point it out. 😓 Or is there some paradoxical property regarding shrinking balls and their elements moving further apart?

• Do you mean the $2$-adic integers, as topological space? For context, see this question. Nov 16, 2022 at 10:57
• I don't understand the problem. Doesn't this map the ball of radius one centered at $x=0$ to the ball of radius $1/2$ centered at $x=-1/3$? Just what you would expect from a shrinking map. Not that $2$-adic balls would have a unique center, but that's not the key issue here. Nov 16, 2022 at 11:13
• But have you not answered the question of homeomorphism yourself? It is clear that $\frac{2x-1}{3}$ is a $p$-adic unit for any $x\in\mathbb{Z}_2$, so it's clearly not surjective. Nov 16, 2022 at 11:45
• You observed that $0$ is not in the image, so it isn't surjective, hence not a homeomorphism. Nov 16, 2022 at 11:46
• @samerivertwice What is your definition of a homeomorphism? The usual one is a continuous bijective function with a continuous inverse. Nov 16, 2022 at 11:48

A homeomorphism needs to biject by definition so $$f$$ trivially isn't a homeomorphism.
Because however, $$f$$ is a homeomorphism onto its image, it is called a topological embedding.