 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-20/6/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Using the transpose formulas of the preceding part,

ε_{FC} F (η_{C}) = α_{C, FC}^{- 1} (α_{C, FC} (1_{FC})) = 1_{FC} .

Similarly,

G (ε_{D}) η_{G D} = α_{G D, D} (α_{G D, D}^{- 1} (1_{G D})) = 1_{G D} .

Thus the adjunction unit and adjunction counit satisfy both triangle identities for an adjunction,

ε_{F} \circ F η = 1_{F}, Gε \circ η_{G} = 1_{G} .

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