By Eugenia L. Cheng

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**Sample text**

Suppose we have a limit cone for D, ( I DI kI DI)I∈މ HkI We need to show that ( (ރ, I DI) (ރ, DI))I∈ މ. is a limit for H• ◦ D. e. HC ◦D. But we have already shown this, since representables preserve limits, and the given cone is just HC of ( I DI DI)I∈ މ. () Suppose F : ފ × މ have a functor D is such that FJ : މ I D has a limit F(I, ) : J such that J I F(I, J) ∼ = I I F(I, J) for all J ∈ ފ. Then we F(I, J) (I,J) F(I, J) in the sense that if one exists, then so does the other, and they are isomorphic with corresponding limit cones.

Free algebras We can define a forgetful functor: U : CT θ (TA A C A) A B f f We may ask two obvious questions: does U have a left adjoint; and does T arise naturally from an adjunction? CT . U has a left adjoint F : C We construct F as follows: • • on objects, FA = on morphisms, F(A T2 A µA TA f , the “free algebra on A”; B) = T2 A µA TA Tf T2 B µB TB . We need to check three things: that FA and Ff satisfy the axioms for an algebra and a map of algebras; that F is functorial; and that F is left adjoint to U.

That K is essentially surjective. So does K hit all of the coequalisers? That is, can we find something in D which goes to each coequaliser? Well, if D has all the “special coequalisers” and G preserves them, then we can lift along U T , so seeing that K sends it to the right place. Hence we get F G is monadic iff D has and G preserves G-very-special coequalisers, and every object of D is a coequaliser of free ones. Can we avoid mentioning free objects in D? In fact, the coequaliser in question is FGY FG Y ; Y and G of this is a coequaliser in C, so it suffices to prove that G reflects these.