I have a field F, with axioms that hold for $+ ,* ,- , /$ and it has a neutral elements $n$ and $e$ . Now, $F1$ is a set and $f:F\rightarrow F1$ a bijective function. I have to show that $F1$ with the following operations: $$x+y := f(f^{-1}(x) + f^{-1}(y))$$ $$x*y := f(f^{-1}(x) * f^{-1}(y))$$ $$-x := (-f^{-1}(x))$$ $$x^{-1} := f((f^{-1}(x))^{-1})$$
with the neutral elements $n1:=f(n)$ and $e1=f(e)$ is also a field.
The way I would solve this would simple be:
Because the fuinction is bijective, I can deduce that $f^{-1}(x) = x$ and $f^{-1}(y)= y$. Since F is a field, the newly defined operations e.g: $x+y :=f(x+y)$ obeys the field axioms and because the codomain is F1 $f(x+y) \in F1$
Any hints how I prove this formally?
Edit: Apparently I started the wrong way.
It is rather tedious to prove all field axioms. To prove commutativity of $+$ you can show that if $x,y\in F_1$: $$x+y=f\big(f^{-1}(x)+f(y)\big)=f\big(f^{-1}(y)+f^{-1}(x)\big)=y+x$$ since addition is commutative between $f^{-1}(x)$ and $f^{-1}(y)$ which are elements of $F$.
To prove distributivity, let $x,y,z\in F_1$ and let us show that $x*(y+z)=x*y+x*z$: $$x*(y+z)=x*f\big(f^{-1}(y)+f^{-1}(z)\big)=f\left(f^{-1}(x)*f^{-1}\Big(f\big(f^{-1}(y)+f^{-1}(z)\big)\Big)\right)$$ by definition. Since $f$ and $f^{-1}$ are inverse of one another, we get $$x*(y+z)=f\left(f^{-1}(x)*\big(f^{-1}(y)+f^{-1}(z)\big)\right)$$ Using distributivity in $F$ we get $$x*(y+z)=f\left(f^{-1}(x)*f^{-1}(y)+f^{-1}(x)*f^{-1}(z)\right)$$ In the mean time $$x*y+x*z=f\left(f^{-1}(x*y)+f^{-1}(x*z)\right)=f\left(f^{-1}(x)*f^{-1}(y)+f^{-1}(x)*f^{-1}(z)\right)$$ which shows that $x*(y+z)=x*y+x*z$.
The other axioms follow the same train of thought.