 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-122/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

By the monoidal coherence theorem, calculations may suppress the canonical associators and unitors, restoring them uniquely afterwards. In this notation the two Frobenius monoidal functor identities read

ψ_{A \otimes B, C} φ_{A, B \otimes C} = (φ_{A, B} \otimes 1) (1 \otimes ψ_{B, C}),

ψ_{A, B \otimes C} φ_{A \otimes B, C} = (1 \otimes φ_{B, C}) (ψ_{A, B} \otimes 1) .

Write

E = ψ_{0} F (e) φ_{X, Y}, N = ψ_{Y, X} F (n) φ_{0} .

FX

, expand and use the first Frobenius identity:

(E \otimes 1) (1 \otimes N) = (ψ_{0} \otimes 1) (F e \otimes 1) (φ_{X, Y} \otimes 1) (1 \otimes ψ_{Y, X}) (1 \otimes F n) (1 \otimes φ_{0}) = (ψ_{0} \otimes 1) (F e \otimes 1) ψ_{X \otimes Y, X} φ_{X, Y \otimes X} (1 \otimes F n) (1 \otimes φ_{0}) = (ψ_{0} \otimes 1) ψ_{I, X} F (e \otimes 1) F (1 \otimes n) φ_{X, I} (1 \otimes φ_{0}) = 1_{FX} .

The third line uses naturality of

ψ

and

φ

. The last line uses

(e \otimes 1) (1 \otimes n) = 1_{X}

, followed by the opmonoidal and monoidal unit axioms.

For

F Y

, the second Frobenius identity gives the other calculation:

(1 \otimes E) (N \otimes 1) = (1 \otimes ψ_{0}) (1 \otimes F e) (1 \otimes φ_{X, Y}) (ψ_{Y, X} \otimes 1) (F n \otimes 1) (φ_{0} \otimes 1) = (1 \otimes ψ_{0}) (1 \otimes F e) ψ_{Y, X \otimes Y} φ_{Y \otimes X, Y} (F n \otimes 1) (φ_{0} \otimes 1) = (1 \otimes ψ_{0}) ψ_{Y, I} F (1 \otimes e) F (n \otimes 1) φ_{I, Y} (φ_{0} \otimes 1) = 1_{F Y} .

Thus both triangular identities hold, and

E : FX \otimes F Y \to I, N : I \to F Y \otimes FX

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