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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?

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  • $\begingroup$ Do you mean the $2$-adic integers, as topological space? For context, see this question. $\endgroup$
    – Dietrich Burde
    Nov 16, 2022 at 10:57
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    $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Nov 16, 2022 at 11:13
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    $\begingroup$ 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. $\endgroup$
    – SomeCallMeTim
    Nov 16, 2022 at 11:45
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    $\begingroup$ You observed that $0$ is not in the image, so it isn't surjective, hence not a homeomorphism. $\endgroup$
    – Jyrki Lahtonen
    Nov 16, 2022 at 11:46
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    $\begingroup$ @samerivertwice What is your definition of a homeomorphism? The usual one is a continuous bijective function with a continuous inverse. $\endgroup$
    – SomeCallMeTim
    Nov 16, 2022 at 11:48

1 Answer 1

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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.

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