I am working through the proof of Theorem 5.3.3 in the following text Deitmar's Automorphic Forms.
The subgroup $\mathbb{Q}^\times$ is discrete in $\mathbb{A}^\times$ and the quotient $\mathbb{A}^1/\mathbb{Q}^\times$ is compact. More precisely, there is a canonical isomorphism $$\mathbb{A}^1/\mathbb{Q}^\times \cong \widehat{\mathbb{Z}}^\times.$$ In particular, $F^1 := \widehat{\mathbb{Z}}^\times \times \{1\}$ is a set of representatives of $\mathbb{A}^1/\mathbb{Q}$ and $ F:= \widehat{\mathbb{Z}}^\times \times (0, \infty)$ is a set of topological representatives of $\mathbb{A}^1/\mathbb{Q}^\times$. The absolute value induces an isomorphism of topological groups: $\mathbb{A}^\times \cong \mathbb{A}^1 \times (0, \infty)$.
To prove the isomorphism $\mathbb{A}^1/\mathbb{Q}^\times \cong \widehat{\mathbb{Z}}^\times$ the author considered the map $\eta : \prod_p \mathbb{Z}_p^\times \to \mathbb{A}^1/\mathbb{Q}^\times$ given by $x \mapsto (x,1) \mathbb{Q}^\times$ and he define the inverse map given by $x := (x_{fin}, x_\infty) \mapsto \frac{1}{x_\infty} x_{fin}$ where he notes that if $x \in \mathbb{A}^1$ then $x_\infty \in \mathbb{Q}^\times$.
My question is why is the inverse map $\eta^{-1} : \mathbb{A}^1/\mathbb{Q}^\times \to \prod_p \mathbb{Z}_p^\times$ given by $x := (x_{fin}, x_\infty) \mapsto \frac{1}{x_\infty} x_{fin}$ even well-defined?
How the author justify that $|\frac{1}{x_\infty} {x_{fin}}_p|_p = 1$ for all finite primes?
Here the notations are $\mathbb{A}^1 := \{x \in \mathbb{A}^\times : |x| = 1\}$ and $\widehat{\mathbb{Z}}^\times := \prod_{p < \infty} \mathbb{Z}_p^\times$. Thank you, any hints/solutions are well-appreciated.
Could I check if there are any gaps in the argument below?
Denote $x_{fin} := (x_2, x_3, x_5,\dots) \in \prod_{p < \infty} \mathbb{Q}_p^\times$ and $$y:= \frac{1}{x_\infty} {x_{fin}} = (y_2, y_3, y_5, \dots) \in \prod_{p < \infty} \mathbb{Q}_p^\times$$ where $y_p = x_p/x_{\infty}$ for each $p< \infty$. Given a finite prime $p < \infty$, write $x_p= p^k u$ with $k \in \mathbb{Z}$ and $u \in \mathbb{Z}_p^\times$ so $|x_p|_p = p^{-k}$ and $$(x_{fin}, 1_\mathbb{Q})|_{\mathbb{A}} = \frac{1}{|x_\infty|_\infty}$$ Then, $$ \frac{1}{|x_\infty|_\infty} = p^{-k} r$$ where $r \in \mathbb{Q}$ so that $r$ is coprime to $p$ in $\mathbb{Q}$. So by computing $|y_p|_p$ we get $$|y_p|_p = \left| \frac{x_p}{x_\infty} \right|_p = \left| \frac{1}{x_\infty} \right|_p |x_p|_p = |p^{-k} r|_p |p^k u|_p = p^k\cdot p^{-k} = 1$$ as desired.