Isomorphism of Banach spaces implies isomorphism of duals?

Solution 1:

For the first part, your argument is okay.

I'd argue:

1. For every Banach space $V$ we have $(1_{V})^\ast = 1_{V^\ast}$
2. For bounded linear maps we have $(ST)^\ast = T^\ast S^\ast$.

This implies that for $S: V \to W$ and $T: W \to V$ such that $ST = 1_{W}$ and $TS= 1_{V}$ that $T^\ast S^\ast = 1_{W^\ast}$ and $S^\ast T^\ast = 1_{V^\ast}$, in other words $(S^{-1})^\ast = (S^\ast)^{-1}$.

As for isometric isomorphisms, check that the inverse of an isometric isomorphism is isometric and that the adjoint of an isometric isomorphism is isometric, too.

No need to invoke the open mapping theorem anywhere.

Solution 2:

Some hints

1) Every functor preserves isomorphisms.

2) The map $$ \begin{align} {}^*&:\operatorname{Ban}\to\operatorname{Ban}&:&W\mapsto W^*\\ &:\mathcal{B}(U,V)\to\mathcal{B}(V^*,U^*)&:&T\mapsto T^* \end{align} $$ is a contravariant functor from category of Banach spaces into category of Banach spaces.

3) For the case of isometric isomorphism consider restriction of $^*$ functor on the category of Banach spaces with contractive maps.