Surjection from decomposition group to automorphism group of residue fields for infinite Galois extensions Let $L/K$ be a (infinite) Galois extension, $A \subset K$ a Dedekind domain with field of fractions $K$ and $B$ its integral closure in $L$. Then for every nonzero prime ideal $\mathfrak{P}$ of $B$ and $\mathfrak{p}$ of $A$, there is a group homomorphism $D(\mathfrak{P}|\mathfrak{p}) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p}))$. I read somewhere that this homomorphism is surjective, and I am trying to deduce this from the case where $L/K$ is finite (which is a standard fact from algebraic number theory). I figured the easiest way to do this deduction would be to write (using the usual arguments)
$$D(\mathfrak{P} | \mathfrak{p}) \cong  \varprojlim_{F \in \mathcal{F}} D_{F/K}(\mathfrak{P} \cap F | \mathfrak{p})$$
and 
$${\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p})) \cong \varprojlim_{F \in \mathcal{F}} {\rm Aut}(\kappa(\mathfrak{P} \cap F)/\kappa(\mathfrak{p}))$$
where $\mathcal{F}$ is the collection of intermediate fields $K \subset F \subset L$ such that $F/K$ is finite and Galois. So the homomorphism we want to prove is surjective is given in the $F$-coordinate by the one in the finite case, 
$$D_{F/K}(\mathfrak{P} \cap F | \mathfrak{p}) \to {\rm Aut}(\kappa(\mathfrak{P} \cap F)/\kappa(\mathfrak{p}))$$
which we know is surjective. But this doesn't seem like enough to deduce the result I want. In particular, pulling back each coordinate under this map will not necessarily yield a valid element of the inverse limit. Is there some way to fix this issue? In particular, I think it should be possible to take $\mathcal{F}$ to be a smaller collection of finite Galois subextensions with the property that for each $F_1, F_2 \in \mathcal{F}$, either $F_1 \subset F_2$ or the other way around. Then, because $I_{F_2/K}(\mathfrak{P} \cap F_2|\mathfrak{p})$ surjects onto $I_{F_1/K}(\mathfrak{P} \cap F_1|\mathfrak{p})$ under the restriction map, we know that if we are given $\sigma_2 \in D_{F_2/K}(\mathfrak{P} \cap F_2|\mathfrak{p})$ that reduces mod $\mathfrak{P} \cap F_2$ to a particular $\tau_2$ which restricts to $\tau_1$ on $\kappa(\mathfrak{P} \cap F_1)$, and $\sigma_1 \in D_{F_1/K}(\mathfrak{P} \cap F_1 | \mathfrak{p})$ reduces mod $\mathfrak{P} \cap F_1$ to $\tau_1$, then we can modify $\sigma_2$ by composing with the appropriate element of $I_{F_2/K}(\mathfrak{P} \cap F_2 | \mathfrak{p})$ so that $\sigma_2$ restricts to $\sigma_1$. The problem is, we can only do this one preimage at a time, and the collection $\mathcal{F}$ is not necessarily countable. Is there some kind of induction/Zorn's lemma argument to fix this? Also, is there a simpler or more direct way to prove this fact?
 A: Pass from $K$ to the fixed field of the decomposition group $L^{D(\mathfrak{P}|\mathfrak{p})}$ so that we may assume $\rm{Gal}(L/K)=D(\mathfrak{P}|\mathfrak{p})$.
Let $I(\mathfrak{P}|\mathfrak p)=\ker(D(\mathfrak{P}|\mathfrak p) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p}))$ be the inertia group and let $L^{I(\mathfrak P|\mathfrak{p})}$ be the inertia field. Then $L':=L^{I(\mathfrak{P}|\mathfrak{p})}$ is Galois over $K$ (as we assumed that  $\rm{Gal}(L/K)=D(\mathfrak{P}|\mathfrak{p})$) and if we let $\mathfrak{P}'=\mathfrak{P} \cap L'$, then we have $\kappa(\mathfrak{P}')=\kappa(\mathfrak{P})$ and the homomorphism $\rm{Gal}(L/K) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p})) $ factors over $\rm{Gal}(L/K) \to \rm{Gal}(L'/K)$ which is surjective. But now $\mathfrak{p}$ is unramified in $L'/K$ which means that we can reduce to the unramified case. But in the unramified case, the map $\rm{Gal}(L'/K) \to {\rm Aut}(\kappa(\mathfrak{P}')/\kappa(\mathfrak{p}))$ is actually an isomorphism, which can be easily concluded from the finite degree case, without the problems that arise if you just try to show surjectivity directly with a limit argument.
