Grp as a reflexive/coreflexive subcategory of Mon So my question is the statement made in the title, is there a functor $F:$ Mon$\to$ Grp which makes Grp into a (co)reflexive subcategory of Mon?
Thanks in advance.
 A: The forgetful functor $U : \mathrm{Grp} \to \mathrm{Mon}$ preserves limits and satisfies the solution set condition, thus has a left adjoint according to the General Adjoint Functor Theorem (see Mac Lane's book for details). You can write it down explicitly: It maps $M$ to the group $G(M)$ which is the free group defined by generators $\underline{m}$, one for each element $m \in M$, subject to the relations $\underline{1}=1$ and $\underline{mn}=\underline{m} \underline{n}$. Thus, elements of $G(M)$ have the form $\underline{m_1} \cdot \underline{m_2}^{-1} \cdot \underline{m_3} \cdot \underline{m_4}^{-1} \cdot \dotsc \cdot \underline{m}_n$. When $M$ is commutative, this is usually called the Grothendieck construction; here every element has the form $\underline{m} \cdot \underline{n}^{-1}$. But as you can see, these kind of adjoint functors always exist when we just forget some part of algebraic structure.
The forgetful functor also has a right adjoint; it maps a monoid $M$ to its group of units $M^*$. This can be verified directly. But there is also a more general approach. Namely, from the construction of colimits in these categories it is clear that $U$ preserves colimits. Again one can verify the solution set condition (for the dual categories), so that $U$ has a right adjoint $R$. The underlying set of $R(M)$ is then $\cong \hom(\mathbb{Z},R(M)) \cong \hom(U(\mathbb{Z}),M) \cong M^*$. Thus, $R(M)=M^*$. Even if you don't know the Adjoint Functor Theorem, this methods lets you to calculate the right adjoint when you don't have a right guess.
In summary, $\mathrm{Grp}$ is a reflective subcategory of $\mathrm{Mon}$, the reflector being the "Grothendieck construction", as well as a coreflective subcategory, the coreflector being the group of units construction.
