Free Monoid is Unique

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Theorem

Let $S$ be a set.

Let $\struct {M, i}$ and $\struct {N, j}$ be free monoids over $S$.


Then there is a unique monoid isomorphism $f: M \to N$ such that:

$\size f \circ i = j$
$\size {f^{-1} } \circ j = i$

where $\size {\, \cdot \,}$ denotes the underlying set functor on $\mathbf{Mon}$.


Proof

By the (categorial) definition of free monoid, we have the following commutative diagram:

$\begin{xy} <-4em,4em>*{\mathbf{Mon} :}, <-4em,1em>*{\mathbf{Set} :}, <0em,4em>*+{N} = "N", <4em,4em>*+{M} = "M", <8em,4em>*+{N} = "N2", "N";"M" **@{.} ?>*@{>} ?*!/_1em/{\bar i}, "M";"N2" **@{.} ?>*@{>} ?*!/_1em/{\bar j}, <0em,1em>*+{\left\vert{N}\right\vert} = "NN", <4em,1em>*+{\left\vert{M}\right\vert} = "MM", <8em,1em>*+{\left\vert{N}\right\vert} = "NN2", <4em,-3em>*+{S} = "S", "NN";"MM" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar i}\right\vert}, "MM";"NN2" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar j}\right\vert}, "S";"NN" **@{-} ?>*@{>} ?*!/_1em/{j}, "S";"MM" **@{-} ?>*@{>} ?<>(.7)*!/_.6em/{i}, "S";"NN2" **@{-} ?>*@{>} ?*!/^1em/{j}, \end{xy}$

The outer rim of these diagrams also needs to satisfy the universal property.

Since obviously $\operatorname{id}_N$ fits in place of $\bar j \circ \bar i$, we conclude these are equal.

Exchanging the rĂ´les of $\struct {M, i}$ and $\struct {N, j}$ proves that $\bar i \circ \bar j = \operatorname{id}_M$ as well.


Hence $\bar i$ is the sought monoid isomorphism.

Its uniqueness follows from the universal property of the free monoid.

$\blacksquare$


Sources