Map from absolute Galois group of a global field to that of one of its completions Let $K$ be a global number/function field, and let $v$ be a place of $K$. How to construct an explicit map from $G(\overline{K}/K)\rightarrow G(\overline{K}_{v}/K_v)$?
 A: The map is defined as follows. It depends on the choice of an embedding $i:\overline{K}\rightarrow\overline{K_v}$ over the canonical $K\rightarrow K_v$ (in the function field case you want the bar to denote separable closure). If you use a different embedding, you will end up conjugating the end result by an element of $\mathrm{Gal}(\overline{K}/K)$. Now, to define $\varphi:\mathrm{Gal}(\overline{K_v}/K_v)\rightarrow\mathrm{Gal}(\overline{K}/K)$, take an element $s$ of the local Galois group. The maps $s\circ i$ and $i$ are both $K$-monomorphisms of $\overline{K}$ into $\overline{K_v}$. Since $\overline{K}$ is normal over $K$, these embeddings have the same image (namely the elements of $\overline{K_v}$ which are separable algebraic over $K$). Thus $\varphi(s):=i^{-1}\circ s\circ i$ makes sense, and is an automorphism of $\overline{K}$ fixing $K$ (alternatively, $\varphi(s)$ is the unique element of $\mathrm{Gal}(\overline{K}/K)$ with $i\circ\varphi(s)=s\circ i$).
You can check from the definition that the homomorphism $\varphi$ so defined is continuous with kernel $\mathrm{Gal}(\overline{K_v}/i(\overline{K})K_v)$. It follows from Krasner's lemma that $i(\overline{K})K_v=\overline{K_v}$, so the kernel is trivial and the map is an injection. Its image is equal to the decomposition group in $\mathrm{Gal}(\overline{K}/K)$ attached to the place of $\overline{K}$ arising from the chosen embedding $i$.
In general, this same procedure can be applied to any extension of fields $E/F$ (choosing separable closures and a compatible embedding). The resulting homomorphism will always be continuous, but it need not be injective.
