 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-18/7/a/solution

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

f : A \to B

is a morphism with an inverse

f^{- 1} : B \to A

satisfying

f^{- 1} f = 1_{A}

and

f f^{- 1} = 1_{B}

With the given one-sided inverses,

g = g 1_{B} = g (f h) = (g f) h = 1_{A} h = h .

Consequently $f$ is an isomorphism with $f^{- 1} = g = h$ . This argument uses only the category axioms.

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