LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Finite words satisfy the free-monoid universal property

Statement

For a set $X$ , let

$X^{*} : = ⋃_{n \in N} X^{n}$

be the set of finite words in letters from $X$ , where $X^{n}$ is the set of functions $n \to X$ . With concatenation and the empty word, $X^{*}$ is a monoid. The one-letter map $i_{X} : X \to X^{*}$ has the universal property that every function $u : X \to M$ into a monoid extends uniquely to a monoid homomorphism $\hat{u} : X^{*} \to M$ .

Facts & Assumptions

Given: A set $X$ , a monoid $(M, \cdot, e)$ , and a function $u : X \to M$ .

[F1]

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

[F2]

A finite product in a monoid is uniquely defined by the recursion $P (0) = e$ and $P (n + 1) = P (n) g_{n}$ (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).

[F3]

The natural numbers form the smallest inductive set (The natural numbers $N$ (von Neumann)).

[F4]

Natural addition satisfies $n + 0 = n$ and $n + (m + 1) = (n + m) + 1$ (Addition of natural numbers).

[F5]

Natural addition is associative: $(n + m) + r = n + (m + r)$ (Addition is associative).

[F6]

For sets $A, B$ , the functions $A \to B$ form a set (The set $B^{A}$ of all functions $A \to B$ ).

[F7]

The indexed union of a family $(A_{i})_{i \in I}$ is $⋃_{i \in I} A_{i} = ⋃ {A_{i} : i \in I}$ ( $⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ , and $⋂_{i \in I} A_{i} : = ⋂ {A_{i} : i \in I}$ for $I \neq \emptyset$ ).

[F8]

If a property holds at $0$ and passes from $n$ to $n + 1$ , it holds for every natural number (The principle of mathematical induction).

Proof

technique · direct

1.1

Each $X^{n}$ is a set by [F6], and [F3] and [F7] make their indexed union $X^{*}$ a set. The unique function $0 \to X$ is the empty word.

F3F6F7

1.2

For $p : n \to X$ and $q : m \to X$ , define $p q : n + m \to X$ by using $p$ on the first $n$ positions and $q$ on the following $m$ positions.

F4construct

1.3

For a word $p : n \to X$ , define $\hat{u} (p)$ as the finite product $u (p (0)) \dots u (p (n - 1))$ , with value $e$ when $n = 0$ .

F2construct

2.1

Function extensionality and [F5] show $(p q) r = p (q r)$ ; the equations in [F4] show that the empty word is a two-sided identity. Thus $X^{*}$ is a monoid by [F1].

step 1.1step 1.2F1F4F5

2.2

Induction on the length of the second word, using [F2] and associativity in $M$ , proves $\hat{u} (p q) = \hat{u} (p) \hat{u} (q)$ . Hence $\hat{u}$ is a monoid homomorphism and extends $u$ on one-letter words.

step 1.2step 1.3F1F2F8

3.1

If $T : X^{*} \to M$ is any homomorphism extending $u$ , induction on word length gives $T (p) = u (p (0)) \dots u (p (n - 1)) = \hat{u} (p)$ ; the base case uses the empty word and the identity of $M$ .

step 2.1step 1.3F1F8

4.1

Thus the extension exists and is unique for every $u$ , including $X = \emptyset$ , where $X^{*}$ contains only the empty word.

step 2.2step 3.1∎

Depends on

Used by

The free-monoid functor is left adjoint to the underlying-set functor Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 47 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Emily Riehl, Category Theory in Context, 2nd ed., Example 4.1.10 (standard reference, not scraped)
Tom Leinster, Basic Category Theory, Example 2.1.3 (standard reference, not scraped)