Solution
= Solution
Using the transpose formulas of the preceding part,
$$
\varepsilon_{FC}F(\eta_C)=\alpha^{-1}_{C,FC}(\alpha_{C,FC}(1_{FC}))=1_{FC}.
$$
Similarly,
$$
G(\varepsilon_D)\eta_{GD}=\alpha_{GD,D}(\alpha^{-1}_{GD,D}(1_{GD}))=1_{GD}.
$$
Thus the <adjunction unit> and <adjunction counit> satisfy \b[both <triangle identities for an adjunction>],
$$
\boxed{\varepsilon_F\circ F\eta=1_F,\qquad G\varepsilon\circ\eta_G=1_G.}
$$