Restriction map between Galois groups induces surjection on inertia groups. Let $M/L/K$ all be algebraic extensions of $\mathbb Q$ and let $P$ be a prime in $\mathcal O_K$, $Q$ be a prime above $P$ in $L$, and $U$ be a prime above $Q$ in $M$. Prove that the map $G(M/K) \twoheadrightarrow G(L/K)$ induces a surjection between the corresponding inertia groups $I(U|P) \longrightarrow I(Q|P)$. Thanks!
Edit:
The fields involved are all arbitrary algebraic extensions of $\mathbb Q$, not necessarily finite.
 A: $\newcommand{\quot}[2]{{\raise{.2em}{#1}\left/\raise{-.2em}{#2}\right.}}$
Here I first assume that the extensions are finite over $\Bbb Q$.
Let $q : G(M/K) \twoheadrightarrow G(L/K)$ be the quotient map (given by the restriction to $L$).
First of all, we easily see that $q(I(U \mid P)) \subset I(Q \mid P)$. Just write the definition of the inertia subgroup.

Secondly, we check that $I(Q \mid P) \subset q(I(U \mid P))$.
The kernel of the morphism 
$$q\vert_{ I(U \mid P) } :  I(U \mid P)  \to  I(Q \mid P)$$
is $I(U \mid Q)$. Therefore the quotient
$$ \quot{I(U \mid P)}{I(U \mid Q)} $$
embeds in $I(Q \mid P)$, and the cardinalities of these groups are respectively: $\dfrac{e(U \mid P)}{e(U \mid Q)}$ and $e(Q \mid P)$, where $e(\cdot \mid \cdot)$ denotes the ramification index. 
But as you know, ramification indices are multiplicative: $e(U \mid P) = e(U \mid Q) e(Q \mid P)$. This implies that the previous embedding is actually an isomorphism, meaning that $q\vert_{ I(U \mid P) }$ is surjective, i.e. $I(Q \mid P) = q(I(U \mid P))$, as desired.

For the general case (with possibly infinite extensions), one can argue using the fact that the Galois groups are profinite groups – the details are explained in these notes by B. Conrad.
