# Are these adjoint functors to/from the category of monoids with semigroup homomorphisms?

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?

• Hi @Thomas, if you think you can answer your own question, then post the answer, so everybody can learn and/or perhaps improve/comment your answer. This is encouraged by StackExchange rules – magma Mar 10 '14 at 9:20
• generally forgetful functors are indicated with "U" (stands for "underlaying") or sometimes "G" (also used for right adjoints in general), while "F" is used for free functors and left adjoints in general. You might consider editing your question in this light. – magma Mar 10 '14 at 9:25
• Are you sure that $G_H$ and $G_O$ are functors? – Martin Brandenburg Mar 10 '14 at 14:48
• @MartinBrandenburg At least I tried to check that $F(g\circ f)=F(g)\circ F(f)$ for $G_H$. (However, I was wrong, as I figured out now.) I didn't even try to check this for $G_O$... It fails, because $g$ might map the identity element to a "false" identity element, so that $G_O(g\circ f)$ would map the newly adjoint identity element to the identity element, but $G_O(g)\circ G_O(f)$ maps the newly adjoint identity element to a "false" identity element. – Thomas Klimpel Mar 10 '14 at 15:42

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.

The well known functor $F: \bf Semigroup \to \bf Monoid$ is the left adjoint to the forgetful functor $G: \bf Monoid \to \bf Semigroup$. To see that $\hom(F(X),Y) \equiv \hom(X,G(Y))$ natural in the variables $X$ and $Y$, explicitly writing down the natural transformations $\varphi$ and $\psi$ inverse to each other helps.

• For each semigroup homomorphism $g:X\to G(Y)$, the monoid homomorphism $\psi(g):F(X)\to Y$ maps the adjoint identity element to the identity element, and is otherwise identical to $g$.
• For each monoid homomorphism $f:F(X)\to Y$, the semigroup homomorphism $\varphi(f):X \to G(Y)$ is the restriction of $f$ to $X$.

The compositions $\varphi\circ \psi$ and $\psi \circ \varphi$ are both identities.

If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_H$ and $G_H$, then $\psi \circ \varphi$ is an identity, but $\varphi\circ \psi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms).

If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_O$ and $G_O$, then $\varphi\circ \psi$ is an identity, but $\psi \circ \varphi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms) when $X$ is not a monoid.

As Martin Brandenburg wrote in a comment, it is not clear whether $F_H$ and $F_O$ are functors at all. In fact, they fail to be functors, because $F(g\circ f)=F(g)\circ F(f)$ fails.

For $F_H$ look at $\begin{matrix} & f & & g & \\ & & * & \longrightarrow & * \\ * & \longrightarrow & * & \longrightarrow & * \\ * & \longrightarrow & * & {}^{\underline{\quad}\nearrow} \end{matrix}$. We get $\begin{matrix} & F_H(f) & & F_H(g) & \\ & & * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} & * & \longrightarrow & * \\ * & \longrightarrow & * & {}^{\underline{\quad}\nearrow} \end{matrix}$, so $\begin{matrix} F_H(g) \circ F_H(f)\\ \begin{matrix} * & \longrightarrow & * \\ * & \longrightarrow & * \end{matrix}\end{matrix}$.

But $\begin{matrix} & g \circ f & * \\ * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} \end{matrix}$, so we get $\begin{matrix} F_H(g \circ f)\\ \begin{matrix} * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} & * \end{matrix}\end{matrix}$. Hence $F_H(g) \circ F_H(f)$ and $F_H(g \circ f)$ are different.

Here, the vertical stars "describe" specific semilattices (commutative idempotent semigroups), where the result of a multiplication is the lower of the two involved stars (i.e. the meet or the greatest lower bound). Hence the identity element is given by the highest star in each case. The arrows "describe" specific homomorphisms.

For $F_O$ look at $\begin{matrix} & f & & g & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * & \longrightarrow & * \end{matrix}$. The leftmost semilattice is missing an identity element, hence $F_O$ will adjoint one. We get $\begin{matrix} & F_O(f) & & F_O(g)\\ * & {}_{\overline{\;\cdot\;}\searrow} & & & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * & \longrightarrow & * \end{matrix}$, so $\begin{matrix} F_O(g) \circ F_O(f) \\ \begin{matrix} * & {}_{\overline{\;\cdot\;}\searrow} & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}\end{matrix}$.

But $\begin{matrix} & g \circ f & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}$, so we get $\begin{matrix} F_O(g \circ f) \\ \begin{matrix} * & \longrightarrow & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}\end{matrix}$. Hence $F_O(g) \circ F_O(f)$ and $F_O(g \circ f)$ are different.