THE MONADICITY THEOREM Consider Two Functors E ⇄ B. I And

THE MONADICITY THEOREM Consider Two Functors E ⇄ B. I And

THE MONADICITY THEOREM BEN ELIAS AND ALEX ELLIS Consider two functors R E B: I ∼ I and R are adjoint if HomE (IM; N) = HomB(M; RN) naturally in M and N. Examples: forget (1) Grp Set. free Res (2) A -mod B -mod for B ⊂ A. Ind f∗ f (3) Shv(Y ) Shv(X) for X Y . f ∗ ∼ Equivalent description: fro all N 2 Ob(E) we have HomB(RN; RN) = Hom(IRN; N). Looking at the image of the identity, we get a counit : IR ! 1E : ∼ Similarly, by looking at the image of the identity under HomE (IM; IM) = HomB(M; RIM) we get the unit η : 1B ! RI: They satisfy certain snake diagrams. Examples: forget (1) Grp Set. In this case IR(G) is the free group on elements free of G. Then the counit IR(G) ! G is given by multiplication. However, there is no functorial map G ! IR(G). Similarly, RI(X) are words in x; x−1. Then the map X ! RI(X) is the inclusion of singletons. Res (2) A -mod B -mod. The counit IR ! 1 is the multiplication Ind A ⊗B A ! A (this is a map of A-bimodules). The unit B ! A is the inclusion (the morphism as B-bimodules). Definition. A monad on C is an endofunctor T : C!C equipped with the multiplication T ◦ T ! T and the unit 1 2 BEN ELIAS AND ALEX ELLIS η : 1 ! T satisfying the associativity and right and left unit axioms. Dually, one can define a comonad. Definition. An algebra over the monad T is an object M 2 C equipped with the action map T ◦ M ! M satisfying the associativity and the unit axioms. The monad of an adjunction: let (I;R) be an adjunction. Here R : E!B and I : B!E. Adjunction gives the counit and unit η. Define a monad in B by T = RI. The unit for the adjunction η gives the unit η for the monad. The multiplication map µ : T ◦ T ! T comes from RIRI ! RI, where we used the counit IR ! 1. One can check that the unit and associativity axioms follow from the snake diagrams for the adjunction. Given a monad (T; η; µ) in C define CT be the category of T -algebras in C and functors IT : C!CT and RT : CT !C. Data of a T -algebra is (X; h), where X 2 C and h is a morphism T x ! x. Then IT (x) = (T X; µx) and RT (x; h) = x. Theorem 1. The functors (IT ;RT ) form an adjunction and the result- ing monad is (T; η; µ). Consider T E B B ; where T is the monad coming from E B. T Theorem 2. Let B B be the adjunction of the previous theorem, where T is the monad of B E. Then there is a unique functor E !BK T , such that RT K = R and KI = IT . K is called the comparison functor. BT is final among all categories which are adjoint to B in a way giving rise to T . Note, there is also the initial category called the Kleisli category BT . Definition. We say R : E!B is monadic if there is an adjunction (I;R) with the corresponding comparison functor an equivalence. Given an object x 2 E we have a map T Rx = RIRx ! RX. This gives the T -algebra structure on Kx. Key trick: TTX ⇒ TX ! X: THE MONADICITY THEOREM 3 The two maps are given either by multiplication in the monad or the action map. We also have the maps X ! TX and TX ! TTX coming from the unit map in the monad. a e Definition. A fork is M ⇒ N ! P with ea = eb. b Forks form a category. Coequalizer is the initial object of that cate- gory (i.e. a colimit). Definition. Coequalizer is absolute if FM ⇒ FN ! FP is coequal- izer for any F . Definition. Coequalizer is split if we have maps P ! N ! M which split the maps. Note, that split coequalizers are absolute. φ Definition. F creates coequalizers if FM FN ! X is a coequalizer, e ⇒ then there is a unique map M ⇒ N ! P , such that F sends it to the previous diagram and P is a coequalizer. Example. Consider M 2 E. Then we have IRIRM ⇒ IRM ! M. Although we have a splitting IRM ! IRIRM, we don't have a map M ! IRM. However, if we apply R we get T T RM ⇒ T RM ! RM, which is split. So, if R creates coequalizers, then M is a coequalizer. Theorem 3 (Barr-Beck). The following are equivalent: (1) E !BK T is an equivalence. (2) E !BR creates coequalizers for (a) absolute (b) split. Proof. 1 ) 2. Just check. 2b ) 1 Construct K−1. If we apply R to IRIRX ⇒ IX, then we get TTX ⇒ TX ! X. −1 Since R is split, we get IRIRX ⇒ IX ! K (X). Exercise: consider the adjunction between semigroups and sets. Un- wind what SetT is and convince yourself that Barr-Beck holds. discrete Non-example: B = Set . Then R is not monadic. forget Suppose f : Y ! X is an open covering, i.e. Y = tiUi. f ∗ ∗ Consider the adjunction Shv(X) Shv(Y )..Then f preservers col- f∗ imits and f∗ preserves limits. 4 BEN ELIAS AND ALEX ELLIS Then the category of descent data is equivalent to T Shv(Y ) (coalge- bras for T ). By the dual Barr-Beck, we get an equivalence T Shv(Y ) ∼= Shv(X)..

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