= Solution
If $B$ belongs to the <full subcategory> $\mathcal C$, the <adjunction> and fullness of its inclusion give, for every $B'\in\mathcal B$,
$$
\boxed{\mathcal B(KB',B)=\mathcal C(KB',B)\cong\mathcal B(B',B),
\qquad h\longmapsto h\rho_{B'}.}
$$
Therefore \b[membership implies the hom-set condition]. In addition, $J$ is <fully faithful>, so the <adjunction counit> is invertible by the preceding criterion. The triangular equation $J\varepsilon_B\,\rho_B=1_B$ then shows that $\rho_B$ is an <isomorphism>. As in the printed inclusion notation, $J$ is suppressed when writing $KB$ as an object of $\mathcal B$.
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