Does $0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$ always split? Let $K$ be a henselian valuation field with residue field $k$, then the decomposition group surjects onto Galois group of the residue field, with kernel the inertia subgroup, namely we have short exact sequence:$$0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$$
When $K$ is a local field, we can split the sequence by lifting the Frobenius; when $K=k((t))$, we can split the sequence by lifting the Galois action (with trivial action on $t$). But in general, do we know if the sequence always split?
(The splitting of the sequence is mentioned as a well-known fact in Proposition A5 in "Exposant et indice d'algèbres simples centrales non ramifiées", but I couldn't find a reference.. Splitting of the sequence would imply that the restriction morphism $H^i_{et}(\mathrm{Spec}(V),M)\to H^i_{et}(\mathrm{Spec}(K),M)$ is injective for any locally constant sheave $M$  on the valuation ring $V$.)
 A: Good question! Let me try to guess what Gabber had in mind there. (Note that he only says "known" (to him), not "well-known"...)
The claim is that the extension splits. Note that to prove this, we are free to replace $K$ by any (algebraic) extension $K'$ whose residue field $k'$ is purely inseparable over $k$. By Zorn's lemma, we can choose $K$ so that it admits no further such extensions. In particular, $K$ is perfect and the value group is divisible. Let $p$ be the characteristic of $k$. Then it follows that $I$ must be pro-$p$, as the maximal unramified extension $K^{\mathrm{ur}}$ of $K$ will still have divisible value group, and so have no nontrivial tame extensions.
Now, if $I$ was nonzero, there is a map $I\to \overline{I}$ where $\overline{I}$ is a nonzero $\mathbb F_p$-vector space (the quotient by the Frattini subgroup). By maximality of $K$, the induced sequence
$$0\to \overline{I}\to \overline{G}\to \mathrm{Gal}_k\to 0$$
is nonsplit, so gives a nonzero class in $H^2(\mathrm{Gal}_k,\overline{I})$. But Galois groups in characteristic $p$ have $p$-cohomological dimension $\leq 1$ by Artin-Schreier theory.
