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 $g$.