Im reading a proof of the fundamental theorem of Galois theory and I'm having trouble understanding the proof of the following part:
Let $F/K$ be a finite Galois extension with $G = \operatorname{Gal}(F/K)$. Then, the function which assigns an intermediate field $E$ the group $\operatorname{Aut}(F/E)$, is a bijection of the set of intermediate fields to the subgroups of $G$, and its inverse sends each subgroup of $G$ to its fixed field.
The proof of injectivity goes as follows:
Let $E$ be an intermediate field and $H = \operatorname{Aut}(F/E)$. Then $H<G$. Let $$E' = F^H = \{a \in F: \forall\sigma\in H, \sigma(a)=a\}$$
Then $E'$ is a field and $E\subset E'\subset F$.
$[E':E]_s = 1$ because if $\psi \colon E' \to \overline{E} = \overline{k}$ is a morphism that is the identity when restricted to $E$, $\psi$ extends to an element in $\operatorname{Aut}(F/E)$, that fixes the elements of $E'$; and so is unique.
In case the notation is weird, $[E':E]_s$ is the number of morphisms $\psi \colon E' \to \overline{E}$ such that they are the identity when restricted to $E$.
I can't understand why is the extension of $\psi$ uniquely determined. Also for context, a Galois extension in this course is one that is separable and normal.
Let $\sigma:F\to F$ be an automorphism that extends $\psi$. Then $\sigma\in H$, so by definition of $E'$, every element of $E'$ is fixed by $\sigma$. So the restriction of $\sigma$ to $E'$, which is just $\psi$, must be the identity map.