Free Monoid is Unique

Theorem
Let $S$ be a set.

Let $\left({M, i}\right)$ and $\left({N, j}\right)$ be free monoids over $S$.

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


 * $\left\vert{f}\right\vert \circ i = j$
 * $\left\vert{f^{-1}}\right\vert \circ j = i$

where $\left\vert{\cdot}\right\vert$ 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 rôles of $\left({M, i}\right)$ and $\left({N, j}\right)$ 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.