Is the free monoid on $n$ generators just the semigroup on $n$ generators adjoined with an identity element?

Question

Is the free monoid on $n$ generators isomorphic to the semigroup on $n$ generators, just adjoined with an identity element? That is, to get the free monoid, you just take the free semigroup, and then add a new element $0$ and stipulate that $0$ is the identity element?

Answer 1

Yes. Here is an abstract way to see this that generalizes to many other situations. The general fact is that adjoints compose: that is, if $F, G$ are two functors with left adjoints $L_F, L_G$, and the composition $F \circ G$ exists, then $F \circ G$ also has a left adjoint and $L_{F \circ G} = L_G \circ L_F$. (And the same for right adjoints, of course.)

So: the forgetful functor from monoids to sets factors as a composite

$$\text{Mon} \to \text{Semi} \to \text{Set}$$

meaning that its left adjoint factors as a composite of left adjoints

$$\text{Set} \to \text{Semi} \to \text{Mon}.$$

The first left adjoint is the free semigroup while the second left adjoint is the free monoid on a semigroup, which is obtained by adjoining an identity. And their composite is the left adjoint of the forgetful functor from monoids to sets, so the free monoid functor.

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Here is a similar but more interesting example of this phenomenon. The forgetful functor from rings to sets factors in (at least) two different interesting ways; first, we can take the factorization

$$\text{Ring} \to \text{Ab} \to \text{Set}$$

where we take the underlying additive group. The left adjoint then factors as a composite

$$\text{Set} \to \text{Ab} \to \text{Ring}$$

where the first functor sends a set $X$ to the free abelian group $\mathbb{Z}[X]$ and the second functor sends an abelian group $A$ to the tensor algebra $T(A)$. This exhibits the free ring on a set $X$ as the tensor algebra $T(\mathbb{Z}[X])$.

The forgetful functor also factors as a composite

$$\text{Ring} \to \text{Mon} \to \text{Set}$$

where this time we take the underlying multiplicative monoid. The left adjoint then factors as a composite

$$\text{Set} \to \text{Mon} \to \text{Ring}$$

where the first functor sends a set $X$ to the free monoid $M_X$ on $X$ and the second functor sends a monoid $M$ to the monoid ring $\mathbb{Z}[M]$. This exhibits the free ring on a set $X$ as the monoid ring $\mathbb{Z}[M_X]$. These are both useful and important ways of understanding the free ring, and very similar remarks apply to the cases of $k$-algebras, commutative rings, and commutative $k$-algebras.