When is the restriction map $Gal(L/F) \rightarrow Gal(K/F)$ a surjection?

I was wondering, when the restriction map $$Gal(L/F) \rightarrow Gal(K/F)$$ is a surjection? I found a good answer to this question here. In point 3), Starfall explains why with the hypothesis of the theorem, the automorphism extends, and I understand that in order to extend the automorphism we need that $$L/F$$ is Galois. But why we need $$K/F$$ to be Galois? I also posted a comment there but maybe I won't get a response.

• If $K/F$ is a normal extension then $\sigma \mapsto \sigma|_K$ in an homomorphism $Gal(L/F) \to Gal(K/F)$ and it is surjective if $L/F$ is normal (thus with kernel $Gal(L/K)$). If $K/F$ is not normal then you need pick first the subgroup $\{ \sigma \in Gal(L/F), \sigma(K) = K\}$ before applying $\sigma \mapsto \sigma|_K$. – reuns Dec 1 '18 at 23:26
• Okay, but I am a little confused. I am reading a book called introduction to abstract algebra by Keith Nicholson, and there is this theorem that says : Let $E/F$ be a Galois extension and let $G=gal(E:F)$. If $K$ is a stable intermediate field, then $K'=gal(E:K)$ is normal in $G$ and $G/K' \cong \{ \lambda \in gal(K:F) \mid \lambda \text{ extends to an automorphism of E} \}$. But since $E/F$ is Galois, don't we even have $G/K' \cong gal(K:F)$ by the arguments you gave? – roi_saumon Dec 2 '18 at 13:07
• Yes, and by the argument given by starfall that any $\sigma\in Aut(K/F)$ extends to automorphisms of the Galois (normal) extensions $L/F$ above $K/F$. For the intuition : in the usual separable setting take a primitive element $K = F(a)$ and look at $L$ the splitting field of $a$'s minimal polynomial, then at $L(b)$. – reuns Dec 2 '18 at 22:12