Beck's theorem

A fork

**Theorem.** Let $Q^*\dashv Q_*$, $Q^*: A\to B$, be an adjoint pair of functors with unit $\eta$ and counit $\epsilon$, and $\Omega:B\to B$ the corresponding comonad. Then $Q^*$ is comonadic, i.e. the comparison functor $K:A\to B^\Omega$, $M\mapsto (Q^* M, Q^*M\stackrel{Q^*(\eta_M)}{\to}Q^*Q_*Q^*M)$ is an equivalence iff $Q^*$ has the property that every fork

in $A$ such that its image by $Q^*$ is a split equalizer sequence in $B$, is itself an equalizer sequence.

We now sketch a proof in the presence of axiom of choice. The fact that the condition on the forks is necessary follows from the fact that the condition is invariant under an equivalence of categories over $B$, while for the Eilenberg-Moore (EM) category the property is easy to check. For the opposite condition, we sketch a slightly nonstandard direct proof, exhibiting a quasiinverse $K^{-1}$ of the comparison map; for the Beck’s theorem in above form this is really easy for objects with a bit more discussion using universality of equalizers for morphisms.

The crucial observation is that, for any $\Omega$-comodule, $(N,\rho)\in B^\Omega$ the diagram

manifestly exhibits a split equalizer sequence; hence, by the assumption, for each $\Omega$-comodule $(N,\rho)$ there is an equalizer of the form

Using the axiom of choice we can therefore form a map $K^{-1}:Ob B^\Omega\to Ob A$. Once a choice of map $K^{-1}$ is made, for any $f:(N,\rho)\to(N',\rho')$, by the universality of equalizers, one has a unique map $K^{-1}(f):K^{-1}(N,\rho)\to K^{-1}(N',\rho')$ for which the diagram

sequentially commutes. Again, by the universality of equalizers, it is easy to show that this rule is functorial; hence $K^{-1}$ becomes a functor.

It is now sufficient to exhibit natural isomorphisms $K K^{-1}\cong Id_{B^\Omega}$ and $K^{-1} K\cong Id_A$.

**Special case of Beck theorem.** Let $Q^* \dashv Q_*$ be an adjoint pair $\mathbf{T}$ its associated monad, and $\mathbf{G}$ its associated comonad.

If $Q_*$ preserves and reflects coequalizers of all parallel pairs in $A$ (for which coequalizers exists) and if any parallel pair mapped by $Q_*$ into a pair having a coequalizer in $B$ has a coequalizer in $A$, then the comparison functor $K : B \rightarrow A^{\mathbf{T}}$ is an *equivalence* of categories.

If $Q^*$ preserves and reflects equalizers of all parallel pairs in $B$ (for which equalizers exists) and if any parallel pair mapped by $Q^*$ into a pair having an equalizer in $A$ has an equalizer in $\B$, then the comparison functor $K' : A \rightarrow \mathbf{G}-{Comod}$ is an equivalence of categories.

See also monadic functor, monadic adjunction.

Revised on March 31, 2014 04:18:13
by Zoran Škoda
(193.136.196.12)