 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-18/2/ii/solution

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

d_{Y}

along

(f, g)

, with maps

e : E \to X

and

k : E \to Y

satisfying

(f, g) e = d_{Y} k

. Applying the two product projections gives

f e = k = g e

, so

e

equalizes

f, g

Given

h : Z \to X

with

f h = g h

, put

v = f h

. Then

(f, g) h = (v, v) = d_{Y} v

. The pullback property gives a unique

u : Z \to E

with

e u = h

and

k u = v

. Any map satisfying

e u = h

automatically has

k u = f e u = f h = v

, so uniqueness is exactly the equalizer uniqueness. Hence pulling back the diagonal constructs the equalizer of $f, g$ .

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