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

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?

• Are you sure [p-adic-number-theory] applies? Nov 16, 2022 at 12:24
• What does $\frac{1}{2} |x-y|_2$ mean? Nov 16, 2022 at 12:26
• @aschepler, why wouldn't that tag apply? Presumably $\Bbb Z_2$ is the $2$-adic integers here, and $\tfrac 12 |x - y|_2$ refers to the $2$-adic norm. Seems like a reasonable tag to me. Nov 16, 2022 at 12:53
• @aschepler yes I'm sure, in fact it's kind of important otherwise the term $\Bbb Z_2$ might be misinterpreted as misleading notation for $\Bbb Z/2\Bbb Z$. Re the metric, it's half the 2-adic metric as Izaam said. Nov 16, 2022 at 13:00

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, $$f(\{x \in K: d(x,c) \le r\} = \{x \in K: d(x,c+b) \le \lvert a\rvert r\}.$$

$$(K, \lvert \cdot \rvert) = (\mathbb Q_2, \lvert \cdot \rvert_2)$$; $$a=\frac23, b=-\frac13$$; $$\lvert a \rvert = \frac12$$.
$$\mathbb Z_2 = \{x \in K: d(x,0) \le 1 \}$$, 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 $$=\{x \in K: d(x,1) \le \frac12 \} = 1+2 \mathbb Z_2 = \mathbb Z_2^\times$$.