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The free word monoid on $X$ represents $M \mapsto S e t (X, U (M))$

Example

Let $X^{*}$ be the set of finite words in the alphabet $X$ , including the empty word $ε$ . Concatenation and $ε$ make $X^{*}$ a monoid, and the one-letter map $j : X \to X^{*}$ makes it the free word monoid on $X$ .

If $M o n$ denotes the locally small category of monoids and unital monoid homomorphisms and $U : M o n \to S e t$ forgets the monoid structure, then $X^{*}$ represents

$M ⟼ S e t (X, U (M)) .$

The representing bijection sends $ϕ : X^{*} \to M$ to $U (ϕ) \circ j$ .

Facts & Assumptions

Given: A set $X$ ; finite words have a length, the empty word has length zero, every nonempty word is uniquely a shorter word followed by one letter, and concatenation joins the two finite lists.

[F1]

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

[F2]

A monoid homomorphism preserves products and the identity; identity maps and composites are monoid homomorphisms (Monoid homomorphism and group homomorphism).

[F3]

A category has associative composition and identity morphisms; it is locally small when each hom-collection is a set (Category, object, morphism, domain, codomain, identity, composition, and hom-collection, Small, locally small, and large categories).

[F4]

A functor preserves identities and composition, and the functions between two fixed sets form a set (Covariant functor, identity functor, composite functor, and contravariant functor, The set $B^{A}$ of all functions $A \to B$ ).

[F5]

The finite product in a monoid has empty value $e$ and appending one factor multiplies it on the right; its splitting law identifies the product along a concatenated list with the product of the first list followed by the product of the second (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either).

[F6]

A property of all finite lengths follows from the zero case and the step obtained by appending one letter (The principle of mathematical induction).

[F7]

A covariant functor is represented by $R$ when it is naturally isomorphic to the hom-functor out of $R$ (Presheaves, covariantly and contravariantly representable functors, and representations).

Verification

technique · constructive

1.1

Concatenation is associative because joining three finite lists gives the same word in either bracketing, and the empty word is a two-sided identity; hence [F1] makes $X^{*}$ a monoid.

givenF1

1.2

Monoids and their unital homomorphisms form a category by [F2] and the function laws. Its hom-collection from $M$ to $N$ is a subclass of the set of functions $U (M) \to U (N)$ , so it is a set and the category is locally small by [F3] and [F4].

F2F3F4

1.3

Given $f : X \to U (M)$ and a word $w = x_{0} \dots x_{n - 1}$ , define $\hat{f} (w) = \prod_{i < n} f (x_{i})$ . In particular, $\hat{f} (ε) = e_{M}$ and $\hat{f} (j (x)) = f (x)$ .

F5construct

2.1

Sending a monoid to its underlying set and a homomorphism to its underlying function preserves identities and composition, so it defines the functor $U$ . Postcomposition therefore makes $M \mapsto S e t (X, U (M))$ a functor.

step 1.2F2F4

2.2

The splitting law in [F5] gives $\hat{f} (w v) = \hat{f} (w) \hat{f} (v)$ for all words $w, v$ ; together with the empty-word equation, this makes $\hat{f} : X^{*} \to M$ a unital monoid homomorphism extending $f$ .

step 1.3F2F5

3.1

If $q : X^{*} \to M$ is a unital monoid homomorphism with $q j = f$ , then $q (ε) = e_{M} = \hat{f} (ε)$ . If $q (w) = \hat{f} (w)$ , then $q (w j (x)) = q (w) f (x) = \hat{f} (w j (x))$ ; induction [F6] proves $q = \hat{f}$ .

step 1.3step 2.2F2F6

4.1

Thus restriction along $j$ and $f \mapsto \hat{f}$ are inverse bijections $M o n (X^{*}, M) ≅ S e t (X, U (M))$ . If $r : M \to N$ is a monoid homomorphism, then both $r \circ \hat{f}$ and $\hat{U (r) \circ f}$ extend $U (r) \circ f$ , so uniqueness in step 3.1 makes them equal; the bijection is natural in $M$ .

step 2.2step 3.1F2

5.1

By [F7], $X^{*}$ is the claimed representing object. The construction also covers $X = \emptyset$ : then $X^{*} = {ε}$ , and there is exactly one unital homomorphism from it to every monoid.

step 2.1step 4.1F7discharge-construct∎

Depends on

Used by

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Sources

David I. Spivak, Category Theory for Scientists, Definition 3.1.1.15 and Proposition 3.1.4.9 (standard reference, not scraped)

The free word monoid on XX represents M↦Set(X,U(M))M↦Set(X,U(M))