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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

Nothing in the library uses this result yet.

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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))