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The free-monoid monad has monoids as its Eilenberg–Moore algebras

Statement

The monad on Set induced by the free-monoid adjunction sends a set X to the set X∗ of finite words, inserts letters as one-letter words, and flattens words of words by concatenation. Its Eilenberg–Moore category is isomorphic over Set to the category of monoids.

Facts & Assumptions

Given: The free-monoid adjunction between sets and monoids.

[L1]

The free-monoid functor sends X to the monoid of finite words X∗ and is left adjoint to the underlying-set functor (The free-monoid functor is left adjoint to the underlying-set functor).

[L2]

Every adjunction induces a monad, whose unit is the adjunction unit and whose multiplication uses the counit (Every adjunction induces a monad on the domain of its left adjoint).

[L3]

A monoid has an associative binary operation and a two-sided identity (Semigroup and monoid).

Proof

technique · direct
1.1L1L2

By [L1]–[L2], the induced endofunctor is X↦X∗, its unit sends a letter to its one-letter word, and its multiplication concatenates a finite word of finite words. A monoid M therefore gives an algebra M∗→M by evaluating each word.

2.1L3step 1.1

Conversely, for an algebra a:X∗→X, define e=a([]) and x⋅y=a([x,y]). The algebra unit law evaluates one-letter words to their letters, and the multiplication law says evaluation is unchanged by first evaluating subwords; applied to empty, two-letter, and three-letter decompositions, it gives the two unit laws and associativity.

3.1L3step 1.1step 2.1∎

An algebra homomorphism commutes with evaluation, hence preserves the empty word and two-letter words and is a monoid homomorphism. Conversely a monoid homomorphism preserves every finite word evaluation, so it is an algebra homomorphism. These identifications are inverse and unchanged on underlying sets.

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