Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 119 4 iii Solution Created 2026-10-03 Updated 2026-10-05
Use , as required by the formula at zero, and composition of functions as multiplication in . The displayed preserves identities, and for ,both sides send zero to zero. Thus is a functor on the one-object category.
For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations and hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at , and are for . The two unit laws are . Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since , so cannot be invertible.
An algebra for a monad is a map with and . The first equation forces ; monotonicity then forces . Thus , which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms satisfy , and this equation holds precisely when : evaluate at zero for necessity, and at both sides equal .
The Kleisli comparison functor takes its only object to this only algebra and sends toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse . It preserves identities and composition by the comparison construction; directly, and . Bijectivity on objects and arrows makes it an isomorphism of categories: