# Proving surjectivity to define inertia group

I'm having trouble understanding the proof of propsoition 2.6.14 of these notes, in particular the part about surjectivity.

Since $$\bar{\sigma}(\theta)$$ is also a root of $$h$$, it is a root of $$\bar{g}$$, and therefore there exists a root $$θ'$$ of $$g$$ such that $$θ'$$ reduces to $$\bar{\sigma}(\theta)$$.

Why is this true? Is every root of $$\bar{g}$$ a reduction of a root of $$g$$?

In this case we have assumed that $$L/K$$ is normal and as $$g$$ is the minimal polynomial of $$\theta$$, it splits into linear factors. $$g$$ has exactly $$deg(g)$$ distinct roots, each of which is a root of $$\bar{g}$$, which has at most $$deg(g)$$ distinct roots. So in this case, any root of $$\bar{g}$$ lifts to a root of $$g$$.
Now, $$h$$ is the minimal polynomial of $$\bar{\theta}$$, and $$\bar{\theta}$$ is a root of $$\bar{g}$$, so $$h$$ divides $$\bar{g}$$. This implies that any root of $$h$$ is a root of $$\bar{g}$$ and can be lifted to a root of $$g$$.