Posted on November 16, 2018
Tags: explanation Categories: category theory

This post is the sequel to the yesterday’s post Algebras compatible with list monad gives rise to a monoid. To illustrate how Eilenberg-Moore category can be constructed with algebras on monads.

A question arises as we learned that an adjunction gives rise to a monad and a comonad, does the converse hold as well? Is there a natural way to get an adjunction from a monad? The answer to this is yes. The intuition is that with monad $$T$$, we have plenty of choices for $$C$$, the point is to find out these two functors $$L:T\to C$$ and $$R:C\to T$$ that satisfies the triangular identity.

I’ll review from horizontal composition of natural transformations; then I’ll review the definition of adjunctions and how adjunctions give rise to a monad; finally, I’ll talk about how a Eilenberg-Moore category can be constructed and why it is left adjoint to a monad category.

## Horizontal composition between natural transformations

Natural transformations, of course, can be composed. Natural transformations can compose in two different ways: horizontally and vertically. Vertical composition is just plain morphism composition with little surprise. Horizontal composition is what gets more interesting. Here I’ll denote the identity natural transformation over $$F$$ by simply $$F$$ when there is no ambiguity.

Let’s see what happens if we compose $$\alpha\circ F$$ in the following diagram (called a “whiskering”, because of its shape):

$\xymatrix { E & & D \ar@/_2pc/[ll]^{G'}="a" \ar@/^2pc/[ll]_{G}="b" & C \ar[l]^{F} \ar @2{->}_{\alpha} "b";"a" }$

Now we take $$a:C$$, and see what we get on $$(\alpha \circ F)_a$$.

$\xymatrix { G'(F a) \\ & F a \ar[ul]^{G'} \ar[dl]_{G} & a\ar[l]^F \\ G(F a) \ar[uu]_{\alpha_{F a}} }$

From the diagram we can see $$(\alpha \circ F)_a = \alpha_{F a}$$. Now let’s see another version of whiskering:

$\xymatrix { E & D \ar[l]^{G} & & C \ar@/_2pc/[ll]^{F'}="a" \ar@/^2pc/[ll]_{F}="b" \ar @2{->}_{\alpha} "b";"a" }$

Similarly, take $$a: C$$, let’s see what we get by composing $$(G \circ \alpha)_a$$.

$\xymatrix { G (F' a) & F' a \ar[dd]_{\alpha_a}\ar[l]^G & \\ & & a \ar[ul]^{F'}\ar[dl]^{F} \\ G (F a)\ar[uu]_{G \alpha_a} & F a \ar[l]^G & \\ }$

Thus $$(G\circ \alpha)_a = G \alpha_a$$, not exactly symmetric to the left whiskering version.

In the rest of the articles, I’ll using the result from this sections a lot and won’t get into the details.

Given two categories $$C$$ and $$D$$, two morphisms that goes between them $$L: D\to C$$, $$R: C\to D$$. We say $$L$$ is left adjoint to $$R$$ when the following condition holds for all $$a:C$$, $$b:D$$.

$C(L b, a) \simeq D(b, R a)$

By replacing $$a$$ with $$L b$$, we get

$C(L b, L b) \simeq D(b, R(L b))$

On the left we have $$id_{L b}$$, by the isomorphism of hom-set, this morphism will select a morphism on the right: $$b \to R (L b)$$. This is an natural transformation called a unit, denoted as $$\eta: 1_D \to R\circ L$$.

Similarly by replacing $$b$$ with $$R a$$, we can get the dual of unit – the counit – on the left of the hom-set isomorphism: $$\epsilon: L \circ R \to 1_C$$.

An adjunction defined by unit and counit has to satisfy the triangle identities, namely:

$\xymatrix { L \ar[r]^{L\circ \eta} \ar@2{-}[rd]^{} & LRL \ar[d]^{\epsilon \circ L} \\ & L \\ }$

and:

$\xymatrix { R \ar[r]^{\eta \circ R} \ar@2{-}[rd]^{} & RLR \ar[d]^{R\circ\epsilon} \\ & R \\ }$

Adjunction defines a very loose equivalence relation between functors $$L$$ and $$R$$. Where one can think of $$L$$ as to introduce some structures, and $$R$$ to remove the structures. A typical example for adjunction is the free-forgetful adjunction.

In last post I mentioned an adjunction can give rise to a monad, here’s how.

Given an adjunction $$L \dashv R$$, where $$L: D\to C$$ and $$R: C\to D$$, we define a functor $$T = R \circ L$$ be the underlying functor of the monad. Let $$\mu = R \circ \epsilon \circ L$$ be the monad multiplication and the monad unit - $$\eta$$ - be the same as the unit for the adjunction.

We first check the types. $$\mu: RLRL \to RL = T^2 \to T$$, check. $$\eta: 1 \to RL = 1 \to T$$, also check. Then we need to check if $$\mu$$ and $$\eta$$ works subjecting to the monad laws.

First the unit laws:

$\xymatrix { T \ar[d]_{T\circ\eta} \ar[r]^{\eta\circ T}\ar@2{-}[rd]^{} & T^2 \ar[d]^{\mu} \\ T^2 \ar[r]^{\mu} & T }$

This law can be derived from triangle identities by left composing the first identity with an $$R$$ and right composing the second identity with an $$L$$. The following diagram will automatically communites thanks to the triangle identities.

$\xymatrix { RL \ar[d]_{RL \circ \eta} \ar[r]^{\eta \circ RL} \ar@2{-}[rd]^{} & RLRL \ar[d]^{R\circ\epsilon\circ L} \\ RLRL \ar[r]_{R\circ\epsilon\circ L} & RL \\ }$

Now the multiplication law (or associativity law):

$\xymatrix { T^3 \ar[r]^{\mu \circ T} \ar[d]_{T\circ\mu} & T^2\ar[d]^\mu \\ T^2 \ar[r]^\mu & T }$

Or,

$\xymatrix { RLRLRL \ar[rr]^{R\circ\epsilon\circ L \circ RL} \ar[dd]_{RL\circ R\circ\epsilon\circ L} & & RLRL\ar[dd]^{R\circ\epsilon\circ L} \\\ & \\ RLRL \ar[rr]^{R\circ\epsilon\circ L} & & RL }$

This is the naturality square of $$\epsilon$$ left composed with $$R$$! Let’s remove the $$R$$ on the left and see what we get:

$\xymatrix { LRLRL \ar[rr]^{\epsilon_{LRL}} \ar[dd]_{LR \circ f} & & LRL\ar[dd]^{f} \\\ & \\ LRL \ar[rr]^{\epsilon_L} & & RL }$

This is exactly the naturality square of $$\epsilon: LR\to1$$. Thus we proved that the monad associativity law holds naturally just because $$\epsilon$$ is a natural transform.

In fact, an adjunction also gives rise to a comonad on $$RL$$, in a very similar fashion, by defining the counit (extract) to be $$\epsilon$$ and $$\delta$$ (duplicate) to be $$L\circ\eta\circ R$$.

Now let’s get back to monad algebra which we discussed extensively in the last post. Last time we proved that an algebra compatible with a list monad is a monoid, it’s a rather surprising finding.

Let’s now revisit the definition of monad compatible algebra. Given a monad $$(T: C \to C, \eta: 1 \to T, \mu: T^2 \to T)$$ and an algebra $$(a: C, \sigma: T a \to a)$$, we define the coherence conditions as follows:

• $$\sigma \circ \eta_a = 1_a$$
• $$\sigma \circ \mu_a = \sigma \circ T \sigma$$

Algebras that satisfy these conditions are compatible with the given monad, and we call them monad algebras.

Not very surprisingly, monad algebras for a monad $$(T, \tau, \mu)$$ do form a category. The objects are $$(a:C, \sigma:T a\to a)$$, and the morphisms are the same morphisms in $$C$$. This category is called monad algebra category, or Eilenberg-Moore category, denoted as $$\textbf{mAlg}_T$$, or $$C^T$$.

Identity morphism on $$(a,\sigma)$$ in $$C^T$$ is just the identity morphism on $$a$$ in $$C$$. Morphism compositions also work the same way as they’re in $$C$$.

I just missed one thing. Before we can claim $$(T a, \mu_a)$$ is a monad algebra, we need to check if this algebra meets the coherence conditions.

First, $$\mu \circ \eta_{T a} = 1_{T a}$$. This one is just the right identity law for monad, so it automatically holds. Then we check $$\mu\circ \mu_{T a}=\mu\circ T \mu$$. It’s just the monad associativity (multiplication) law! See how these all fits so perfectly. Just because of monad laws, $$\mu_a$$ will always be a compatible algebra on $$T a$$.

The cannonical functor from $$C^T$$ to $$C$$ is the forgetful functor that “forget” about the algebra part, i.e. $$U: (a,\sigma) \mapsto a$$. In addition, $$U: f \mapsto f$$ for morphisms.

It’s much more tricker to define the free functor $$F: C\to C^T$$. We need to find a valid algebra for every $$a$$. Fortunately, it turns out we already have a very good candidate – $$\mu_a: T (T a) \to T a$$ from the monad. $$\mu$$ is a natural transformation so it works on any $$a$$, for every $$a$$ we have an algebra from $$T (T a)$$ to $$T a$$.

Now we can define the free functor $$F: C \to C^T$$ as $$a \mapsto (T a, \mu_a)$$, which is kind of neat. Of course, $$F$$ should also maps $$f: a \to b$$ to $$F f: T a \to T b$$, which is just $$T f$$.

Now we have got a pair of free and forgetful functors, it’s time to prove they are really adjoint. We say $$F$$ is left adjoint to $$U$$, or $$F \dashv U$$.

To play with adjunction, we need to first define our pair of natural transformations $$\eta: 1_C \to U\circ F$$ and $$\epsilon: F\circ U \to 1_{C^T}$$.

Let’s write down $$\eta$$ in its components form: $$\eta_a: a \to U (F a)$$, and we know $$U (F a) = U (T a, \sigma) = T a$$. We can just use the unit $$\eta$$ from the monad!

What about $$\epsilon$$? $$\epsilon_{(a,\sigma)}: F (U (a,\sigma)) \to (a,\sigma)$$. Where $$F (U (a, \sigma)) = F a = (T a, \mu_a)$$. We need to find a map from $$(T a, \mu_a)$$ to $$(a, \sigma)$$. We can just use the forgotten evaluator $$\sigma: T a\to a$$.

In order for the unit and counit to form an adjunction, we need to check the triangle laws.

$\xymatrix { (T a, \mu_a) \ar[r]^{F\circ \eta_a} \ar@2{-}[rd]^{} & (T(T a), \mu_{T a}) \ar[d]^{\mu_{a}} \\ & (T a, \mu_a) \\ }$

We can check the algebras’ carrier types.

$\xymatrix { T a \ar[r]^\eta & T (T a) \ar[r]^\mu & T a }$

This is essentially just $$\mu \circ \eta$$, which is equal to the identity by the right unit law on monad. The evaluator follows automatically, because they’re just regular morphisms in $$C$$. Then we check another triangle identity:

$\xymatrix { a \ar[r]^{\eta_{U (a, \sigma_a)}} \ar@2{-}[rd]^{} & T a \ar[d]^{U\circ\sigma_a} \\ & a \\ }$

Omitting the unimportant part, we get:

$\xymatrix { a \ar[r]^{\eta_a} & T a \ar[r]^{\sigma_a} & a }$

Looking familiar? Yes, $$\sigma \circ \eta = 1$$ must hold by one of the coherence conditions for $$\sigma$$ to be compatible with our monad.

Therefore we have shown the Eilenberg-Moore category $$C^T$$ is left adjoint to a monad category $$C$$. I’m not sure if it’s valid to say a category is left adjoint to another category. But anyway, we have discovered another free-forgetful functor pair that are adjoint to each other, and what makes it so fascinating is that it’s an adjunction we can get from ANY monad.

Given the monad algebra adjunction above, let $$T'=U\circ F$$ be the endofunctor of a monad. On objects, $$T' a = (U\circ F) a = (T a, \mu_a) = T a$$; on morphisms, $$T' f = (U\circ F) f = U (T f) = T f$$. The unit $$\eta'$$ for the new monad is defined to be the unit for the adjunction, which is in turn defined by the same unit on the original monad. The multiplication $$\mu'$$ is defined as $$U\circ \epsilon \circ F$$ from the adjunction, where $$\epsilon$$ is just the the evaluator part for $$(T a, \mu_a)$$, i.e. $$\mu$$ from the same monad!