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There's a well known theorem due to Beck that characterizes when an adjunction is monadic, that is, if $F$ is left adjoint to $G$, $G:D \to C$, $GF:=T$ is always a monad on $C$, and the adjunction is called monadic, essentially, when $D$ is the Eilenberg–Moore category $C^T$ of $T$-algebras and $G$ is the forgetful functor. (For the precise definition see I was wondering if there was a similar characterization to determine when $D$ is the Kleisli category of FREE $T$-algebras?

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up vote 9 down vote accepted

There is a unique functor $\mathbf{Kl}(GF) \rightarrow \mathbf{D}$ commuting with the adjunctions from $\mathbf{C}$, since the Kleisli category is initial among adjunctions inducing the given monad; and this functor is always full and faithful, since $\mathbf{Kl}(GF)(A,B) \cong \mathbf{C}(A,GFB) \cong \mathbf{D}(FA,FB)$.

So this functor will be an equivalence iff it is essentially surjective, and an isomorphism iff it is bijective on objects. But its object map is just the object map of $F$.

So $\mathbf{Kl}(FG)$ is equivalent to $\mathbf{D}$ compatibly with the adjunctions from $\mathbf{C}$ precisely when $F$ is essentially surjective, and isomorphic just when $F$ is bijective on objects.

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Thanks very much! – David Carchedi May 27 '10 at 6:21
So, I guess this implies that if $F$ is left adjoint to $G$ and $G$ does not reflect isos, then $F$ cannot be essentially surjective? – David Carchedi May 27 '10 at 6:28
Yep, I think so! More generally, $G$ will always be full and faithful on the essential image of $F$, and hence reflect isomorphisms there. – Peter LeFanu Lumsdaine May 27 '10 at 15:34

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