Beck’s Monadicity Theorem

Now we have introduced the idea of monadicity, we move onto the most important theorem in this area. Beck’s monadicity theorem tells us that a functor $U: \mathcal{B} \rightarrow \mathcal{C}$ is monadic if and only if the following three conditions hold:

U has a left adjoint.
U reflects isomorphisms.
$\mathcal{B}$ has coequalizers of reflexive $U$ -contractible pairs, and $U$ preserves them.

The aim of this post is to unpack these three conditions, and to justify why they are at least necessary for a monadic functor. That they are also sufficient is a rather amazing fact that we may delve into in another post. The first two conditions are relatively trivial, but the third will require more consideration.

Left adjoints

The first condition, that $U$ has a left adjoint is the most trivial. By definition, we require a monadic functor to be a right adjoint so we can consider the comparison functor to the Eilenberg-Moore category.

Isomorphism reflection

A functor $U$ is said to reflect isomorphisms, or to be conservative, if for every $f$ , $U(f)$ is an isomorphism implies $f$ is.

Remark: Isomorphism reflection is a statement about morphisms not objects. It does not say that if $U(A)$ is isomorphic to $U(B)$ then $A$ is isomorphic to $B$ .

For a monad $\mathbb{T} : \mathcal{C} \rightarrow \mathcal{C}$ , if we have an Eilenberg-Moore algebra morphism $h : (A,\alpha) \rightarrow (B,\beta)$ , then in $\mathcal{C}$

$h \cdot \alpha = \beta \cdot \mathbb{T}(h)$ .

Now assume there exists $\mathcal{C}$ -morphism $g : B \rightarrow A$ such that

$g \cdot h = \mathsf{id}_{A}$ and $h \cdot g = \mathsf{id}_{B}$

then we would like to show $g$ is an algebra morphism of type $(B,\beta) \rightarrow (A,\alpha)$ . This is a straightforward calculation:

$g \cdot \beta = g \cdot \beta \cdot \mathbb{T}(h \cdot g) = g \cdot h \cdot \alpha \cdot \mathbb{T}(g) = \alpha \cdot \mathbb{T}(g)$ .

Therefore $U^{\mathbb{T}}$ reflects isomorphisms. It is easy to verify that equivalences are also conservative, and conservative functors are closed under composition. As

$U = U^{\mathbb{T}} \circ K$

where $K$ is the comparison functor, if $U$ is monadic, it must be conservative. This condition is already sufficient to weed out various putative monadic functors.

Example: Let $\mathsf{Pos}$ be the category of posets and monotone maps, and $U : \mathsf{Pos} \rightarrow \mathsf{Set}$ the obvious forgetful functor. This functor has a left adjoint, and so satisfies the first condition of the monadicity theorem. Now consider the two element posets:

The poset $X$ , with underlying set $\{ x_1, x_2 \}$ with the discrete partial order.
The poset $Y$ , with underlying set $\{ y_1, y_2 \}$ with the least partial order such that $y_1 \leq y_2$ .

There is a bijective monotone map $h : X \rightarrow Y$ with $h(x_1) = y_1$ and $h(x_2) = y_2$ . This morphism has no inverse monotone map. On the other hand, the underlying function $U(h)$ has an obvious inverse in $\mathsf{Set}$ . Therefore $U$ is not conservative, and therefore cannot be monadic.

This argument is easily adapted to show the forgetful functor $\mathsf{Cat} \rightarrow \mathsf{Set}$ taking a small category to its set of objects is not monadic.

The coequalizer condition

We now need to untangle the final condition, and this will be a little bit more technical. We need to introduce two related notions.

Firstly, a parallel pair:

$d^0, d^1 : A \rightarrow B$

is said to be contractible (or split) if there exists $t : B \rightarrow A$ such that

$d^0 \cdot t = \mathsf{id}_B$ and $d^1 \cdot t \cdot d^0 = d^1 \cdot t \cdot d^1$ .

Notice that these conditions are asymmetrical in $d^0$ and $d^1$ .

Secondly, a contractible (or split) coequalizer consists of the following data:

A parallel pair $d^0, d^1 : A \rightarrow B$ .
A morphism $t : B \rightarrow A$ .
A morphism $d : B \rightarrow C$ .
A morphism $s : C \rightarrow B$ .

such that:

$d$ coequalizes the parallel pair, that is $d \cdot d^0 = d \cdot d^1$ .
$d^0$ has section $t$ , that is $d^0 \cdot t = \mathsf{id}_B$ .
$d$ has section $s$ , that is $d \cdot s = \mathsf{id}_C$ .
$d^1 \cdot t = s \cdot d$ .

Again, notice the asymmetry of the conditions. As suggested by the terminology, a contractible coequalizer is a coequalizer in the ordinary sense.

Furthermore, a contractible coequalizer is simply a collection of morphisms satisfying four equations. If we apply a functor to the morphisms, the resulting data will also satisfy the same equations, as functors preserve them, and will therefore also form a contractible coequalizer. Therefore contractible coequalizers are preserved by every functor. That is, they are an example of an absolute colimit.

A second useful fact, connecting these two notions, is that if a contractible pair $d^0, d^1 : A \rightarrow B$ has a coequalizer $d : B \rightarrow C$ , then there automatically exists an $s : C \rightarrow B$ such that it forms a contractible coequalizer.

A $U$ -contractible pair, is a pair $d^0, d^1 : A \rightarrow B$ such that $U(d^0), U(d^1) : U(A) \rightarrow U(B)$ has a contractible coequalizer. Notice this terminology is slightly misleading, we are requiring a contractible coequalizer, not just a contractible pair.

With that mass of terminology in place, we recall that for any monad $\mathbb{T}$ , the forgetful functor $U^{\mathbb{T}} : \mathcal{C}^{\mathbb{T}} \rightarrow \mathcal{C}$ creates colimits which exist in the base category and are preserved by $\mathbb{T}$ and $\mathbb{T}^2$ . $U$ -contractible pairs induce absolute coequalizers in the base category, which are therefore created by $U^{\mathbb{T}}$ . Secondly, although equivalences don’t necessarily create colimits that exist, they do preserve them. Therefore, if $U : \mathcal{B} \rightarrow \mathcal{C}$ is monadic, as

$U = U^{\mathbb{T}} \circ K$

$\mathcal{B}$ has coequalizers of $U$ -contractible pairs, and $U$ preserves them.

In fact, this is stronger than the proof of the monadicity theorem needs, and we can restrict our attention to a small class of coequalizers. A parallel pair $d^0, d^1 : A \rightarrow B$ is said to be reflexive if they have common section. That is, there exists an $r : B \rightarrow A$ such that

$d^0 \cdot r = \mathsf{id}_B = d^1 \cdot r$ .

Conclusion

Two out of the three conditions of the monadicity theorem are pretty routine. The third took a bit of unravelling, but enduring the technicalities is a worthwhile investment, as these special colimits arise again and again in the theory of monads. We will look in detail at the most important source of these special coequalizers in a later post.

Further reading: Our account owes a lot to that of Barr and Well’s in Toposes Triples and Theories. We have deliberately used similar notational choices to them in case readers want to consult their work for further details, which is highly recommended. Johnstone remarks in the Elephant that the restriction to reflexive pairs in condition three is omitted in many accounts, but is all that is needed. Toposes, Triples and Theories is also careful in this regard.

Beck’s Monadicity Theorem

Left adjoints

Isomorphism reflection

The coequalizer condition

Conclusion

2 thoughts on “Beck’s Monadicity Theorem”

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Left adjoints

Isomorphism reflection

The coequalizer condition

Conclusion

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2 thoughts on “Beck’s Monadicity Theorem”

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