 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-23/4/b/solution

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

p i = 1_{A}

give

(i p) i = i

, so

i

equalizes

i p

and

1_{B}

. If

h : X \to B

satisfies

(i p) h = h

, put

k = p h : X \to A

. Then

ik = i p h = h

. If

k^{'}

is another such factorization, applying

p

gives

k^{'} = p i k^{'} = p h = k

. This proves

i : A \to B is the equaliser of i p and 1_{B} .

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