Restriction from subgroup of the Galois group of max. unr, ext. $G(\tilde{K}/\mathbb{Q}_{p})$ to $G(K/\mathbb{Q}_{p})$ is surjective? This is a question I'm struggling with for some time.
Let $K$ be a finite Galois extension of $\mathbb{Q}_{p}$ and let $\tilde{K}$ denote the maximal unramified extension of $K$.
We can then consider the continuous surjective restriction map of Galois groups
\begin{equation}
G(\tilde{K}/\mathbb{Q}_{p})\to G(K/\mathbb{Q}_{p}).
\end{equation}
I was told that we can actually restrict the domain of this map to the subgroup $(\varphi)$ of $G(\tilde{K}/\mathbb{Q}_{p})$ generated by the Frobenius element $\varphi$ of $K$ and this homomorphism still end up being surjective.
Now, why is that? I strongly feel that this has something to do with the fact that since $G(\tilde{K}/\mathbb{Q}_{p})$ is a procyclic group, since it has $(\varphi)$ as a dense cyclic subgroup , we could pick a pre-image of a given $\sigma\in G(K/\mathbb{Q}_{p})$ as something that could be "approximated" by elements in $G(\tilde{K}/\mathbb{Q}_{p})$, but this is just not precise. Is this some general property of procyclic groups? Thanks.
 A: The reason Frobenius is enough is just that your field is Henselian. Recall that finite, unramified extensions of local fields are determined completely by their residue fields. But then, this being a finite extension of finite fields, the residue field extension Galois group is generated by the Frobenius automorphism. Now when you pass to the profinite limit, the density of $\langle\varphi\rangle$ and the continuity gives you enough to work with just the Frobenius when you restrict.
So in short:  it has some to do with the group being pro-cyclic, but the real key here is that unramified extensions have Galois groups corresponding to the Galois groups of the residue field extension.
A: [This is supposed to be a comment, but I don't know how to type TEX in a comment]
The setting of the question is not clear at all. At the beginning, $K/\mathbf Q_p$ is a Galois, say of group $G$. By maximality of $\tilde {K}$ , the extension $\tilde {K} /\mathbf Q_p $ is Galois, and we have indeed a natural surjective  map $\pi$ from $G(\tilde {K} /\mathbf Q_p)$ onto $G $. But  its restriction to $(\phi)$ must have cyclic image, hence  can surely not be surjective if $G$ is not cyclic. And even if $G$ were cyclic, ramification could come into play in $ K /\mathbf Q_p$ . Suppose that $ K /\mathbf Q_p$ is ramified of degree $p$, hence totally ramified. Then the extension $\tilde {K} /\mathbf Q_p $ would split as the composite of $K$ and $\tilde {\mathbf Q_p}$, and $\pi$ would be an isomorphism of $G(\tilde {K} /K)$ onto $G(\tilde {\mathbf Q_p} /\mathbf Q_p)$. Finally, for the question to make sense, we need $ K /\mathbf Q_p$ to be unramified, but then $\tilde {K}= \tilde {\mathbf Q_p}$ and the answer becomes obvious.
