The setup of the question is the same As the unanswered question here. Here is the relevant part of Neukirch.
Here $G$ is an arbitrary profinite group (but I take it as the Galois group of some extension) and $d : G -> \widehat{\mathbb{Z}}$ is surjective
My question is about the definition of $d_K$ itself. If $f_K$ is finite, then $d(G_K)$ has to be the unique subgroup corresponding to the fixed field of $\mathbb{F}_{p^f}$ and corresponds to those elements of $\widehat{\mathbb{Z}}$ that are $0$ on projecting to $\mathbb{Z}/f\mathbb{Z}$. However the definition of $d_K$, as I understand it, involves computing the $1/f$th element of each $\mathbb{Z}/m\mathbb{Z}$. But then how would this be surjective, as you can't solve $7x = 3$ modulo 7.
Can someone give me the explicit description of this $d_K$ map and explain why it should be surjective?
You can embed $\widehat{\mathbb{Z}}=\prod_p \mathbb{Z}_p$ in $\prod_p \mathbb{Q}_p$, in which every integer $n\ne0$ is invertible, allowing you to divide by $f_K$.
By the definition of $f_K$, we have $d(G_K)=f_k\widehat{\mathbb{Z}}$, so $\frac{1}{f_k}d(G_K)=\widehat{\mathbb{Z}}$.