One claim of the Galois theorem I'm working with is the following:
For $T \leq U$, $U$ a Galois extension, there exists an anti-isomorphism between the lattice of intermediate fields $V$, ($V$ such that $T \leq V \leq U$), and the lattice of subgroups of $Gal(U,T)$.
As the maps, we are using $Gal(U,V)$ - all automorphisms of $U$ that fix $V$, and $Fix(U,G)$ - the field of all elements that every $g$ fixes.
We can use the concept of the Galois correspondence etc. to show that these maps give us our wanted anti-isomorphism. I understand all the steps, besides this - we need to show that $Gal(U,\cdot)$ and $Fix(U, \cdot)$ are onto.
Because for finite $G$ it's true that $Gal(U,Fix(U,G))=G$, we see $Gal(U,\cdot)$ is onto. We also know that $Fix(U,Gal(U,V))=V$ for $V$ a Galois extension of $T$ - but not all the fields in question have to be normal (Galois theorem makes an additional claim that the normal fields correspond to normal subgroups) - therefore I'm not sure how to show that $Fix(U,\cdot)$ is onto.
Why is $Fix(U,\cdot)$ onto?