For example the class of monoids is an equational theory for $$\mathbf{E}=\bigl\{(1\cdot x,\, x),\; (x\cdot 1,\, x),\; \bigl((x\cdot y)\cdot z,\, x\cdot (y\cdot z)\bigr)\bigr\}$$ i.e. all the algebras with two operations: one of arity 0 (the unit) and one of arity 2 (the multiplication), such that the $1$ is the unit for multiplication $\cdot $ and multiplication is associative. The class of commutative monoids is also an equational theory with one additional equation $(x\cdot y,\, y\cdot x)$. Groups, Boolean or Heyting algebras, lattices are also equational theories.

Coming back to free algebras: it turns out that the set of terms $\mathsf{T}^{(f_i)}(X)$ on a given set of variables $X$ has an algebra structure of type $(f_i)_{i=1,\dots,n}$: it is given by the inductive step in the definition of terms: if $t_i\in \mathsf{T}^{(f_i)}(X)$ for $i=1,\dots,j_i$ then $$ f_j^{\underline{\mathsf{T}^{(f_i)}(X)}}(t_1,\ldots,t_{j_i}) := f_j(t_1,\ldots,t_{j_i})\in \mathsf{T}(X) $$ Furthermore $\underline{\mathsf{T}^{(f_i)}(X)}$ is a free algebra over $X$ in the class of all algebras of the given type $(f_i)_{i=1,\dots,n}$. An extension of a map $h:X\rightarrow\underline{A}=(A,(f_i^{\underline{A}})_{i=1,\ldots,n})$ can be build inductively following the definition of terms and using the homomorphism property: $$ h(f_i(t_1,\ldots,f_{i_j})) := f_i^{\underline{A}}(h(t_1),\ldots,h(t_{i_j})) $$ The map $h$ is indeed a homomorphism: $$ \begin{array}{ll} h\bigl(f_i^{\underline{\mathsf{T}(X)}}(t_1,\ldots,t_{i_j})\bigr) & = h(f_i(t_1,\ldots, t_{i_j}) \\\\ & = f_i^{\underline{A}}(h(t_1),\ldots, h(t_{i_j})) \\\\ \end{array} $$ Note that the class of algebras of the same type is usually very broad, but this is the first approximation to build free algebras in an equational theory. This is just the equational theory for the empty set $\mathbf{E}$.

Let’s see this on an example and let us consider algebras of the same type as a monoid: with one nullary operation (unit $1$ or `mempty` if you like) and one 2-ary operation (multiplication / `mappend`). Let $X$ be a set of variables. Then $1$ is a valid term, and also if $t_1$ and $t_2$ are terms on $X$ then also $t_1\cdot t_2$ is a term, but also $t_1\cdot 1$ and $1\cdot t_2$ are valid and distinct terms. $\mathsf{T}(X)$ resembles a monoid but it isn't. It is not associative and the unitality condition is not valid since $t\cdot 1\neq t\neq 1\cdot t$ as terms. We still need a way to enforce the laws. But note that if you have a map $f:X\rightarrow M$ to a monoid $M$ which you'd like to extend to a homomorphism $\mathsf{T}(X)\rightarrow M$ that preserves $1$ (which is not the unit, yet) and multiplication (even though it is not associative), you don’t have much choice: $\mathsf{T}(X)\rightarrow M$: $t_1\cdot t_2$ must be mapped to $f(t_1)\cdot f(t_2)\in M$.

We need a tool to enforce term equations. For that one can use

Equivalence relations and congruences form complete lattices (partial ordered which have all suprema and minima, also infinite). If you have two equivalence relations (congruences) then their intersection (as subsets of $A^2$) is an equivalence relation (congruence).

The set of equations that defines the class of monoids generates a congruence relation on the term algebra $\underline{\mathsf{T}^{f_i}(X)}$ (i.e. an equivalence relation which is compatible with operations: $x_1\sim y_1$ and $x_2\sim y_2$ then $(x_1\cdot y_1) \sim (x_2\cdot y_2)$). One can define it as the smallest congruence relation which contains the set $\mathbf{E}$. Equivalence relation on a set $A$ is just a subset of the cartesian product $A\times A$ (which satisfy certain axioms), so it all fits together! One can describe this congruence more precisely, but we'll be happy with the fact that it exists. To show that, first one need to observe that intersection of congruences is a congruence, then the smallest congruence containing the set $\mathbf{E}$ is an intersection of all congruences that contain $\mathbf{E}$. This intersection is non empty since the set $A\times A$ is itself a congruence relation.

The key point now is that if we take the term algebra and take a quotient by the smallest congruence that contains all the pairs of terms which belong to the set $\mathbf{E}$ we will obtain a free algebra in the equational class defined by $\mathbf{E}$. We will leave the proof to a curious reader.

Free monoids

Let’s take a look on a free monoid that we can build this way. First let us consider the free algebra $\underline{\mathsf{T}(X)}$ for algebras of the same type as monoids (which include non associative monoids, which unit does not behave like a unit). And let $\sim$ be the smallest relation (congruence) that enforces $\mathsf{T}(X)/\sim$ to be a monoid.

Since monoids are associative every element in $\underline{\mathsf{T}(X)}/\sim$ can be represented as $x_1\cdot( x_2\cdot (x_3\cdot\ldots \cdot x_n))$ (where we group brackets to the right). Multiplication of $x_1\cdot( x_2\cdot (x_3\cdot\ldots \cdot x_n))$ and $y_1\cdot( y_2\cdot (y_3\cdot\ldots \cdot y_m))$ is just $x_1\cdot (x_2\cdot (x_3\cdot\ldots\cdot(x_n\cdot (y_1\cdot (y_2\cdot (y_3\cdot\ldots\;\cdot y_m)\ldots)$. In Haskell if you’d represent the set $X$ as a type $a$ then the free monoid is just the list type $[a]$ with multiplication: list concatenation and unit element: the empty list. Just think of

Free Monads

It turns out that monads in $\mathcal{Hask}$ are also an equational theory. Just the terms are higher kinded: $*\rightarrow*$ rather than $*$ as in monoids. The same construction of a free algebra works in the land of monads, but we need to look at them from another perspective. Let us first take a mathematical definition of view on monads.

Definition Monad

A monad is an (endo) functor `m` with two natural transformations:

class Monad m where
return :: a -> m a
join   :: m(m a) -> m a

which is unital and associative, i.e. the following law holds:

-- | associativity
join . join == join . fmap join
-- | unitality
join . return  = id = join . fmap return

These axioms are easier to understand as diagrams:

and

It is a basic lemma that this definition a monad is equivalent to what we are used to in Haskell:

Having `join` one defines `>>=` as

and the other way, having `>>=` then

Monoids in monoidal categories