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 ∈ 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 ∈
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 C T be the category of T -algebras
in C and functors I T : C → C T and RT : C T → C.
Data of a T -algebra is (X, h), where X ∈ C and h is a morphism
T x → x. Then I T (x) = (T X, µx) and RT (x, h) = x.
Theorem 1. The functors (I T , RT ) form an adjunction and the resulting monad is (T, η, µ).
Consider
E B BT ,
where T is the monad coming from E B.
Theorem 2. Let B B T be the adjunction of the previous theorem,
where T is the monad of B E. Then there is a unique functor
K
E → B T , such that RT K = R and KI = I T .
K is called the comparison functor.
B T 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 ∈ E we have a map T Rx = RIRx → RX. This
gives the T -algebra structure on Kx.
Key trick:
T T X ⇒ T X → 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 → T X and T X → T T X 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 category (i.e. a colimit).
Definition. Coequalizer is absolute if F M ⇒ F N → F P is coequalizer 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 F M ⇒ F N → 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 ∈ 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:
K
(1) E → B T is an equivalence.
R
(2) E → B 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 T T X ⇒ T X → X.
Since R is split, we get IRIRX ⇒ IX → K −1 (X).
Exercise: consider the adjunction between semigroups and sets. Unwind 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 = ti Ui .
f∗
Consider the adjunction Shv(X) Shv(Y )..Then f ∗ preservers colf∗
imits and f∗ preserves limits.
4
BEN ELIAS AND ALEX ELLIS
Then the category of descent data is equivalent to T Shv(Y ) (coalgebras for T ). By the dual Barr-Beck, we get an equivalence T Shv(Y ) ∼
=
Shv(X).