I'm trying to prove the following result
A banach space is reflexive if and only if its dual is reflexive
in the most category theoretical way I can.
I managed to prove one implication in a not too categorical way and am now stuck with the other one. I'd appreciate any help with dealing with this problem, either for solving the other implication, either for solving both at once or either for solving the first one in an even more categorical way.
My proof:
We have a contravariant functor $(-)^*:\textbf{Ban}\to \textbf{Ban}$ that assigns to every Banach space its dual, and to every morphism (Bounded linear operator) $T:E\to F$ its transpose $T^*:F^*\to E^*$.
If we let $J_X$ be the canonical embedding of $X$ in its dual $X^*$ $(x\mapsto ev_x)$. The hypothesis that for a Banach space $E$, $J_E:E\to E^{**}$ is a Banach space isomorphism means it's an isomorphism in the category $\textbf{Ban}$, thus by applying the functor, we get an isomorphism $J_E^*:E^{***}\to E^*$ such that for every $\psi\in E^{***}$ and $x\in E$ $$J_E^*(\psi)(x)=\psi(J_E(x))$$
and there exists an isomorphism $\Phi:E^*\to E^{***}$ such that $J_E^*\; \Phi = id_{E^*}$.
Since we also have that for every $f\in E^*$ and $x\in E$ $$J_E^*J_{E^*}(f)(x)=J_{E}^*(ev_f)(x)=ev_f(J_E(x))=ev_f(ev_x)=f(x)$$ then $J_E^*J_{E^*}(f)=f$ which yields $J_E^*J_{E^*}=id_{E^*}$.
We deduce that $\Phi=J_{E^*}$ and thus, $E^*$ is reflexive since $J_E$ is an isomorphism.