If we know that there is a (topological) isomorphism between two Banach spaces $X,Y$ called $\phi \in L(X,Y)$. Then the appropriate isomorphism between the dual spaces $X',Y'$ is given by $\phi' \in L(Y',X')$.
I was wondering: What is the fastest way to see that this $ \phi'$ is actually an isomorphism? Does anybody here have a good idea to show this fast? I mean sure, you could show that it is onto and injective, but this does not seem to be a fast idea.
We can easily construct an inverse to $\phi^*$, it is given by $(\phi^*)^{-1}=(\phi^{-1})^*$, indeed we have: $$\left<(\phi^{-1})^*\phi^*v,x\right>=\left<\phi^*v,\phi^{-1}(x)\right>=\left<v,\phi\phi^{-1}(x)\right>=\left<v,x\right>$$ for all $x\in X$ and $v\in X^*$. Thus $(\phi^{-1})^*\phi^*=1$, and similarly $\phi^*(\phi^{-1})^*=1$.