(For context, $x_1, ... , x_n$ refer to the roots of an $n$th degree polynomial)
On this text (page 4) about Galois theory, the author claims that
"for each radical extension $E$ of $ℚ(x_1, ..., x_n)$ there is a radical extension $E'/E$ with automorphisms extending all permutations $σ$ of $x_1, ..., x_n$."
My understanding of what the author is saying is that after adjoining some radicals into $ℚ$ one could continue to adjoin more different radicals until a field $E'$ with the property of having at least one automorphism for each permutation $σ$ of $x_1, ..., x_n$ is formed.
Is my understanding correct? If not, what does the author mean by "with automorphisms extending all permutations $σ$ of $x_1, ..., x_n$"?
If it is correct, then I don't get why this theorem holds. The proof the author gives is the following:
"for each adjoined element, represented by the expression $e(x_1, ... , x_n)$, and each permutation $σ$ of $x_1, ..., x_n$ adjoin the element $e(σx_1, ... , σx_n)$. Since there are only finitely many permutations of $σ$ then, the resulting field $E'⊇E$ is also a radical extension of $ℚ(x_1, ..., x_n)$.
This gives a bijection (also called $σ$) of $E'$ sending each $f(x_1, ... , x_n)∈E'$ (a rational function of $x_1, ... x_n$ and the adjoined elements) to $f(σx_1, ... , σx_n)$, and this bijection is clearly an automorphism of $E'$, extending the permutation $σ$."
I understand that $E'$ is a radical extension of $ℚ(x_1, ..., x_n)$, is just the automorphism claim that is confusing me. I don't see, for example, why it has to be necessarily a bijection. How could one prove that it is impossible for a property of a permutation $σ$ to make two elements in $ℚ(x_1, ..., x_n)$ to be mapped into the same place?