TITLE: Are these adjoint functors to/from the category of monoids with semigroup homomorphisms?
QUESTION [8 upvotes]: Do the forgetful functors $G_H:\bf Monoid \to \bf Semigroup^1$ and $G_O:\bf Semigroup^1 \to \bf Semigroup$ have left and/or right adjoints? Here $\bf Semigroup$ is the category of semigroups with semigroup homomorphisms, $\bf Semigroup^1$ is the category of monoids (semigroups with identity) with semigroup homomorphisms, and $\bf Monoid$ is the category of monoids with monoid homomorphisms (the image of the identity is the identity).
An interesting functor $F_H:\bf Semigroup^1 \to \bf Monoid$ is given by the functor which leaves the monoid objects unchanged, and modifies the homomorphisms by replacing the image of the identity by the identity of the target monoid. Is this a left or right adjoint functor to $G_H$?
A nice functor $F_O:\bf Semigroup \to \bf Semigroup^1$ is the one which leaves monoids unchanged, and adjoins an identity element to semigroups missing an identity element. The homomorphisms are extended by mapping the newly adoint identity element to the identity element of the target monoid. Is this a left or a right adjoint functor to $G_O$, or both?
A well known functor $F:\bf Semigroup \to \bf Monoid$ is the one which adjoins to each semigroup an identity (independent of whether it already has one), and extends the homomorphisms by mapping the newly adoint identity element to the newly adjoint identity element of the target monoid. I have seen the assumption (statement) that this is the left adjoint of the forgetful functor. Is this correct?

REPLY [3 votes]: Functor $F\circ G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Monoid}$
serves as left adjoint of functor $G_{H}:\mathbf{Monoid}\rightarrow\mathbf{Semigroup}^{\mathbf{1}}$.
Does functor $G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Semigroup}$
have a left adjoint? Let's say it has and let's denote it as $K:\mathbf{Semigroup}\rightarrow\mathbf{Semigroup}^{\mathbf{1}}$.
Then the composition $F\circ G_{O}\circ K:\mathbf{Semigroup}\rightarrow\mathbf{Monoid}$
should serve as left adjoint of $G=G_{O}\circ G_{H}:\mathbf{Monoid}\rightarrow\mathbf{Semigroup}$.
This according to the rule $\left(G_{O}\circ G_{H}\right)^{ad}=G_{H}^{ad}\circ G_{O}^{ad}$.
Then $F$ and $F\circ G_{O}\circ K$ must be isomorphic functors (both
left adjoints of $G$). However, if you start with an object $S\in\mathbf{Semigroup}$
having no identity and a finite underlying set, then $K\left(S\right)$
must have at least $1$ element more (the missing identity must be
'added') and $F\circ G_{O}\left(K\left(S\right)\right)=F\left(K\left(S\right)\right)$
will on its turn have $1$ element more than $K\left(S\right)$ (again
an identiy is added). Then $F\circ G_{O}\left(K\left(S\right)\right)=F\left(K\left(S\right)\right)$
has at least $2$ elements more than $S$ in contrast with $F\left(S\right)$
that has exactly $1$ element more. This shows that the functors cannot
be isomorphic. The conclusion is that $G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Semigroup}$
has no left adjoint.