Notations:
1.$E/F$ is normal,$Gal(E/F)=Aut_F(E)$;
2.$E/F$ algebraic,$Hom_F(E,\bar F)$ is the set of morphisms $\Phi:E\to\bar F$ fixing $F$.
Question:
I met this question in proving the statement that if $K/F$ Galois(assume that $K\subset \bar F$) and $E$ intermediate,$E/F$ Galois,then $Gal(K/F)/Gal(K/E)\simeq Gal(E/F)$.
I worked out a 'proof' without using the separability:
Step 1: if $K/F$ normal(assume that $K\subset \bar F$) and $E$ intermediate,then the restriction map $res_E:Gal(K/F)/Gal(K/E)\to Hom_F(E,\bar F),\sigma Gal(K/E)\mapsto\sigma_E$ is a bijection.
proof of step 1:
$res_E$ is well-defined:$\forall\sigma\in Gal(K/F),\forall\tau\in Gal(K/E),res_E(\sigma\tau)=(\sigma\tau)_E=\sigma_E=res_E(\sigma)$;
$res_E$ is injective:$\forall res_E(\sigma_1)=res_E(\sigma_2),\sigma_2^{-1}\sigma_1\in Gal(K/E)$,so $\sigma_1Gal(K/E)=\sigma_2Gal(K/E)$;
$res_E$ is surjective:$\forall\sigma'\in Hom_F(E,\bar F)$,we can lift it to $\sigma:K\to\bar F$(Thm:$E/F$ algebraic,$\phi:F\to\bar F$ field embedding,there exists $\Phi:E\to\bar F$ extending $\phi$).Since $K/F$ normal,so $\sigma(K)=K$,$\sigma\in Gal(K/F),res_E(\sigma)=\sigma'$.
Step 2: if $E/F$ normal,then this is an isomorphism.
proof of step 2:
Since $E/F$ is normal,so every $\phi:E\to\bar F$ fixing $F$ satisfies $\phi(E)=E$,so $Hom_F(E,\bar F)=Gal(E/F)$;
Map $Gal(K/F)\to Gal(E/F)$ sending $\sigma$ to $\sigma_E$ is a group morphism.It has kernel $Gal(K/E)$.By step 1,$res_E$ is an isomorphism.
Did I use the separability implicitly?Is my proof correct,or does there exist some counterexamples to the weaker statement(both $K/F$ and $E/F$ are merely normal)?Thank you!