Proposition 7

With the same notations as previously, suppose that $d_{\Phi}$ is bijective. Then every homomorphism $u$ of $E$ into $E'$ admits an everywhere defined left adjoint, and one has $u^* = (d_{\Phi})^{-1} \circ {}^t u \circ d_{\Phi'}$.

Indeed, since $d_{\Phi}$ is bijective, one has, with the notations of the beginning of the No., $F'_1 = F'$, and $u^*$ is therefore everywhere defined. On the other hand (30) is equivalent to

now $\langle d_{\Phi'}(y'), u(x) \rangle = \langle {}^t u(d_{\Phi'}(y')), x \rangle$; hence one has ${}^t u(d_{\Phi'}(y')) = d_{\Phi}(u^*(y'))$ for all $y' \in F'$, whence ${}^t u \circ d_{\Phi'} = d_{\Phi} \circ u^*$, and consequently the announced expression for $u^*$. Q.E.D.