= Solution
For $\phi\in V'$ and $v\in V$,
$$
(I'\phi)(v)=\phi(Iv)=\phi(v),
$$
so $I'$ is the <identity map> on the <dual space>. If $S:U\to V$ and $T:V\to W$, then, for $\psi\in W'$ and $u\in U$,
$$
\bigl((TS)'\psi\bigr)(u)=\psi(T(Su))
=\bigl(S'(T'\psi)\bigr)(u).
$$
Equality on all vectors proves
$$
\boxed{(TS)'=S'T'.}
$$
Apply this to $RR^{-1}=R^{-1}R=I$. It gives both $R'(R^{-1})'=I'$ and $(R^{-1})'R'=I'$. Consequently the <dual map> is invertible and
$$
\boxed{(R')^{-1}=(R^{-1})'.}
$$
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