 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-4/1c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Suppose

(g \circ f) (a) = (g \circ f) (a^{'})

g

implies

f (a) = f (a^{'})

, and the injectivity of

f

then implies

a = a^{'}

. Hence

g \circ f

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