 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/1e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

This is always true. If

f (x_{1}) = f (x_{2})

, then

(g \circ f) (x_{1}) = g (f (x_{1})) = g (f (x_{2})) = (g \circ f) (x_{2}) .

The assumed injectivity of

g \circ f

gives

x_{1} = x_{2}

, so

f

is injective.

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