$\DeclareMathOperator{\GL}{GL}\DeclareMathOperator{\ord}{ord}\DeclareMathOperator{\SL}{SL}$I have a question about a proposition from Daniel Bump's book, Automorphic Forms and Representations. Here $A = \mathbb A$ is the adeles of $F$, and $A_f = \prod\limits_{v < \infty}' F_v$ are the finite adeles.
I am trying to understand why (ii) is true in the case $K_0 = \prod\limits_{v < \infty} \GL_n(\mathcal O_v)$. If $\operatorname{Id}(F)$ is the group of fractional ideals of $F$, and $P(F)$ is the group of principal ideals, I want to say that the surjective homomorphism $$\GL_n(\mathbb A) \rightarrow \operatorname{Id}(F)/P(F)$$ $$ x \mapsto \prod\limits_{\mathfrak p < \infty} \mathfrak p^{\ord_{\mathfrak p}(\det(x_{\mathfrak p}))}P(F)$$ induces a bijection of sets $$\GL_n(F) \GL_n(F_{\infty})\backslash \GL_n(\mathbb A)/K_0 \rightarrow \operatorname{Id}(F)/P(F)$$
I've checked that this map is well defined and surjective. For injectivity, I suppose $g, g' \in \GL_n(\mathbb A_f)$ (I may assume from the beginning that $g,g' \in \GL_n(\mathbb A)$ each have trivial infinite part, since if not then I can modify by multiplication from $\GL(F_{\infty})$) with
$$\prod\limits_{\mathfrak p}\mathfrak p^{\ord_{\mathfrak p}(\det(g_{\mathfrak p}))} \equiv \prod\limits_{\mathfrak p}\mathfrak p^{\ord_{\mathfrak p}(\det(g'_{\mathfrak p}))} $$
Then there must exist an $\alpha \in \GL_n(F)$ such that for each (finite) prime $\mathfrak p$,
$$\ord_{\mathfrak p}(\det(\alpha g_{\mathfrak p})) = \ord_{\mathfrak p}(\det( g'_{\mathfrak p})) $$
Then for each prime $\mathfrak p$, $\det(\alpha g_{\mathfrak p})$ and $\det( g'_{\mathfrak p})$ generate the same fractional ideal in $F_{\mathfrak p}$, so there exists a $k_{\mathfrak p} \in \GL_n(\mathcal O_{\mathfrak p})$ such that $\det(\alpha g_{\mathfrak p}k_{\mathfrak p})$ and $\det( g'_{\mathfrak p})$. If we set $k = (k_{\mathfrak p}) \in K_0$, then we have shown that
$$\alpha g k g'^{-1} \in \operatorname{SL}_n(\mathbb A)$$
This doesn't quite give me what I want. I need to further multiply $g$ on the left by $\GL_n(F)$ and on the right by $K_0$ to make it actually equal to $g'$. Any idea?
One way I thought was to let $j$ be the embedding $\SL_n(F) \rightarrow \SL_n(\mathbb A_f)$. By (i), the image of $j$ is dense in $\SL_n(\mathbb A_f)$. Also, $K_0 \cap \SL_n(\mathbb A_f)$ is open in $\SL_n(\mathbb A_f)$, so the product set
$$g'^{-1}j\big(\SL_n(F) \big) \alpha g k\bigg( K_0 \cap \SL_n(\mathbb A_f) \bigg) \subset \SL_n(\mathbb A_f)$$
being open and dense, must equal to $\SL_n(\mathbb A_f)$. So there exists a $\beta \in \SL_n(F)$ and $k' \in K_0 \cap \SL_n(\mathbb A_f)$ such that
$$g'^{-1} \beta^{-1} \alpha gkk'^{-1} = 1$$
which implies $\alpha gk = \beta g'k'$. Unfortunately, this is circular since apparently what I want to prove is used to prove (i).