Inverse of isomorphism between two fields is isomorphism

Question

I'm using this definition of isomorphism between two systems satisfying the axioms of Dedekind-complete totally ordered fields to show that its inverse is an isomorphism too. I can be easily proved that it's a bijection and both of morphism properties characteristic to ordered field isomorphisms. I have problem proving the last condition, that $f(a)\,{\preceq}_Y\,f(b){\implies}f^{-1}(f(a))\,{\preceq}_X\,f^{-1}(f(b))$.

We know by the provided definition that ${\forall}a,b\,{\in}\,X:a\,{\preceq}_X\,b{\implies}f(a)\,{\preceq}_Y\,f(b)$. The usual proof of the condidition mentioned before uses the property $a\,{\preceq}_X\,b{\iff}f(a)\,{\preceq}_Y\,f(b)$, which is indeed often used in definitions of order isomorphisms.

I can't do this, since my source uses a mere ${\implies}$ instead of more convenient ${\iff}$; how can I deduce from my axioms that $a\,{\preceq}_X\,b{\iff}f(a)\,{\preceq}_Y\,f(b)$? If I can't, how to prove that $f(a)\,{\preceq}_Y\,f(b){\implies}f^{-1}(f(a))\,{\preceq}_X\,f^{-1}(f(b))$?

Answer 1

Use the fact that your orderings are total. If $a\not\preceq_X b$, then $b\prec_X a$. This implies $f(b)\preceq_Y f(a)$ and $f(b)\neq f(a)$ since $f$ is injective, so $f(a)\not\preceq_Y f(b)$.

Answer 2

If the fields are totally ordered then,
if $a \nleq b$ then $b < a$, whence $f(b) \leq f(a)$;
in fact, $f(b) < f(a)$, because $f$ is injective;
then $f(a) \nleq f(b)$.

Answer 3

The trick here is that trichotomy partitions everything into three disjoint cases preserved by morphisms:

• $a < b \implies f(a) < f(b)$
• $a = b \implies f(a) = f(b)$
• $a > b \implies f(a) > f(b)$

So having $f(a)$ and $f(b)$ fall into one of the three cases can only happen if $a$ and $b$ are in the same case.