Is there a monoid with universal property dual to that of a free monoid?

Question

One of the important properties of a free monoid on a set is that for any other monoid, you can create a unique homomorphism from the free monoid to that monoid.

Does there exist some monoid over a set that can do the converse? (eg. a monoid $A$ such that for any other monoid $B$ you can create a unique homomorphism from $B$ to $A$)

I know this wouldn't exactly be an inverse free monoid by definition, but it was the closest way I could think of phrasing it.

Answer 1

A free monoid on the set $X$ is a monoid $M$ together with a set theoretic map $i\colon X \to M$ such that

For every monoid $N$ and every set map $j\colon X\to N$, there exists a unique monoid morphism $f\colon M\to N$ such that $j=f\circ i$.

You seem to want a monoid with the dual property. In category theory we would call it a "co-free monoid on $X$."

Two possible ways to interpret this occur to me:

1. Reverse only the final arrow. A co-free monoid on $X$ is a monoid $M$ and a set map $i\colon X\to M$ such that for every monoid $N$ and set map $j\colon X\to N$, there exists a unique monoid morphism $f\colon N\to M$ such that $i=f\circ j$. I think this is what you want.

2. Reverse all arrows. A monoid $M$ together with a set map $i\colon M\to X$ such that for every monoid $N$ and every set map $j\colon N\to X$ there exists a unique monoid morphism $f\colon N\to M$ such that $j = i\circ f$.

For the first case, I claim the only "co-free monoid on $X$" is the trivial monoid with the trivial map $i$. Indeed, take $N$ to be the trivial monoid and $j$ the constant map. Then $f$ must send $e_N$ to $e_M$ (since it is a monoid morphism) so $i=f\circ j$ means that $i$ sends every element of $X$ to $e_M$. Now take $M$ to play the role of $N$, and $i$ to play the role of $j$. Then both the trivial map $M\to M$ and the identity map satisfy the necessary condition $i=f\circ j$, so by the uniqueness clause the two maps are equal and $M$ is the trivial monoid. It is now easy to see that the trivial monoid satisfies the desired property, as there is only one map from $N$ to $M$ and it satisfies $i=f\circ j$ for every $j$.

In the second case I claim that such a monoid exists if and only if $X$ is a singleton, and you just get the trivial monoid again.

Clearly $X$ is nonempty. Let $x=i(e_M)$. If there is any other element $y\in X$, $y\neq x$, then take $N$ to be the trivial monoid, $j$ to map $e_N$ to $y$. Then no monoid morphism $f\colon N\to M$ can satisfy $j=i\circ f$. Thus, $X$ must be a singleton. Now take $M$ and $i$ to play the roles of $N$ and $j$, and note that both the trivial map and the identity satisy $i=i\circ f$. So the uniqueness clause yields that $M$ is the trivial monoid once again. This monoid and map again satisfy the desired property.

So in the first interpretation, you get the trivial monoid for every set. In the second interpretation, it only works if $X$ is a singleton, and again $M$ must be the trivial monoid.