I am studying complete local fields. For the third time i read proofs which seem similar.
There is an application defined on a complete group with a filtration $F_n$ (i'm not sure if it's the right term). I mean there is a sequence of sub-groups $F_{n+1}\subset F_{n}$.
The induced application on quotients $F_n/F_{n+1}$ has a property (onto or iso) then one deduce the application has the same property.
I found that for: norm is onto in unramified extensions, for $x\rightarrow x^q$ is onto in units group (where $F_n=1+\frak p^n$) and $x\rightarrow x^q-x$ in ${\frak o}_L$ ( with $F_n=\frak p^n$).
I think it could exist a general result for such situation that could be used three times.
I'm aware this is a bit unclear but context are quite different and yet proofs looks similar and use completion : One write the property for class in the quotient then construct a inductive sequence that "push" an annoying term to 1 or 0 depending on the law + or x .
Does it seems familiar to someone
This is indeed a general result: Let $G$ and $H$ come with normal filtrations $(G_n)_{n\geq 0}, (H_n)_{n\geq 0}$, such that $G=G_0$, $G_{n+1}\subseteq G_n$ and $\bigcap_{g\geq 0}G_n=0$ and likewise for $H$. Suppose then that we have a homomorphism $\phi:G\to H$ that respects this filtration in that $\phi(G_n)\subseteq H_n$. Then, $\phi$ induces maps on each subquotient $$\phi_n:G_n/G_{n+1}\to H_n/H_{n+1}$$ and we have the following results:
(1) If each $\phi_n$ is injective, then so is $\phi$.
(2) If $G$ is complete and each $\phi_n$ is surjective, then so is $\phi$.
The proofs are not too difficult: For (1), suppose that $\phi(g)=1$ for some $g\in G$. Inductively it is then clear that $g\in G_n$ for each $n$ - but that intersection is trivial so that $g=1$ all along. (Note that this makes no use of completeness.)
As for (2), suppose we have some $h\in H$. Inductively, we can construct a sequence $(g_n)_{n\geq 0}$ such that $g_n\equiv g_{n+1}\mod G_n$ and $\phi(g_n)\equiv h\mod H_n$ for all $n$: For $n=0$, we may just take $g=1$. If $g_n$ is already set, the difference $k=\phi(g_n)h^{-1}$ lies in $H_n$ by assumption. Then, since $\phi_n$ is onto, we may find some $g'\in G_n$ such that $\phi(g')\equiv k \mod H_{n+1}$. Then $g_{n+1}=g'g_n$ works, as $$\phi(g'g_n)h^{-1}\equiv 1\mod H_{n+1}$$ and $g'g_n\equiv g_n \mod G_n$.
As $G$ is complete, the sequence $(g_n)$ must converge to some $g\in G$ with the property $g\equiv g_n\mod G_n$. As a consequence we have \begin{align} \phi(g)h^{-1}&=\underbrace{\phi(gg_n^{-1})}_{\in H_n}\phi(g_n)h^{-1}\in H_n \end{align} for each $n$. But as this intersection is empty, we must have $\phi(g)=h$, so that $\phi$ is also surjective.
For this second implication, the completeness assumption is really essential, as (for instance) the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Z}_p$ induces isomorphisms on the subquotients $p^n\mathbb{Z}/p^{n+1}\mathbb{Z}\simeq p^n\mathbb{Z}_p/p^{n+1}\mathbb{Z}_p$, but is of course not itself surjective.