Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 20 10 iii Solution 2026-10-14
For , consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: satisfies the monad associativity law, while is the image under of . The map is a monad algebra morphism because , and the same equation says it equalizes the pair.
Suppose is a monad algebra morphism with . Put . Naturality of and the unit law giveMoreover, since is a monad algebra morphism,Thus is a monad algebra morphism and factors through . If , then , proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.
The underlying fork is a split coequalizer: take , and . Then , , and . Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in .
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 18 6 c Solution Created 2026-10-03 Updated 2026-10-06
For an algebra for a monad , consider the fork in the Eilenberg-Moore categoryHere is the counit at , and . The arrow is an algebra morphism by , which also says that it coequalizes the two arrows.
Let be an algebra morphism with . Define . Naturality of at givesSince is an algebra morphism,Thus is an algebra morphism with . Any other such factorization satisfies . This proves the full coequalizer universal property inside the algebra category.
The pair is moreover a reflexive pair: its common section is , with underlying map , because and . Hence every monad algebra is a reflexive coequalizer of free algebras. This reflexive free-algebra presentation of a monad algebra needs no general existence theorem for arbitrary algebra-category colimits.