Is $f(x)=(2x-1)/3:\Bbb Z_2\to\Bbb Z_2^\times$ a near-homeomorphism? Semihomeomorphism? Homeomorphism?

Question

Question

Is $f(x)=(2x-1)/3:\Bbb Z_2\to\Bbb Z_2^\times$ a near-homeomorphism? Semihomeomorphism? Homeomorphism?

My Attempt

I have that $f(x)=(2x-1)/3:\Bbb Z_2\to\Bbb Z_2^\times$ is continuous and has a continous inverse because $\lvert f(x)-f(y)\rvert_2=\frac12\lvert x-y\rvert_2$

I have that $f$ can't be a homeomorphism $\Bbb Z_2\to\Bbb Z_2$ because it doesn't surject.

What about $\Bbb Z_2\to\Bbb Z_2^\times$ though? Which is the question.

Looks to me like it's sufficient to prove $f,f^{-1}$ both inject. Both are functions so have an injective inverse, it just remains to prove they land inside the domain and range.

$f^{-1}(\Bbb Z_2^\times)\subseteq\Bbb Z$ because $\lvert f^{-1}(x)\rvert_2\leq0$

and

$f(\Bbb Z_2)\subseteq\Bbb Z_2^\times$ because $\lvert(2x-1)/3\rvert_2=1$

So it's a homeomorphism from $\Bbb Z_2\to\Bbb Z_2^\times$, correct?

Answer 1

Linear maps on valued fields

Let $(K , \lvert \cdot \rvert)$ be any field with an absolute value (which induces the metric $d(x,y):= \lvert x-y \rvert$).

Let $a,b \in K, a \neq 0$.

The map $f:K \rightarrow K, f(x)=ax+b$ has inverse $f^{-1}(x) = \frac{1}{a}x - \frac{b}{a}$.

We have $d(f(x),f(y)) = \lvert a \rvert \cdot d(x,y)$ for any $x,y \in K$. (In particular, if $\lvert a \rvert <1$, then $f$ is a contraction mapping with Lipschitz constant $\lvert a \rvert$.)

For any subset $S \subseteq K$, the restriction of $f$ to $S$ induces a homeomorphism from $S$ to its image $f(S)$. (Since in general $S \neq f(S)$, these restrictions are technically no longer contraction mappings.)

More precisely, for any $c \in K$ and $r \in [0, \infty)$, the restriction of $f$ to $$\{x \in K: d(x,c) < r\}$$ (the "open" ball of radius $r$ around $c$) has image $$\{x \in K: d(x,c+b) < \lvert a \rvert r\}$$ (the "open" ball of radius $\lvert a \rvert r$ around $c+b$). Analogously for "closed" balls, $$f(\{x \in K: d(x,c) \le r\} = \{x \in K: d(x,c+b) \le \lvert a\rvert r\}.$$

Your application

$(K, \lvert \cdot \rvert) = (\mathbb Q_2, \lvert \cdot \rvert_2)$; $a=\frac23, b=-\frac13$; $\lvert a \rvert = \frac12$.

$\mathbb Z_2 = \{x \in K: d(x,0) \le 1 \}$, the "closed" unit ball.

$f(\mathbb Z_2) = \{x \in K: d(x,-\frac13) \le \frac12 \}$.

Now since this value is non-archimedan, hence the metric is an ultrametric, any point inside a ball serves as its centre, and $d(1,-\frac13) = \frac14$, this image is $=\{x \in K: d(x,1) \le \frac12 \} = 1+2 \mathbb Z_2 = \mathbb Z_2^\times$.