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

Question

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?

Answer 1

A homeomorphism needs to biject by definition so $f$ trivially isn't a homeomorphism.

And it contracts in the sense that the smallest ball in which one can contain the range is smaller than the smallest ball containing the domain.

Because however, $f$ is a homeomorphism onto its image, it is called a topological embedding.