How to understand the Grothendieck group as an adjoint?

Question

My main goal is to understand the Grothendieck group construction and its generalization to non-commutative setting. I understand its explicit construction via an equivalence relation, but I want to fully understand as an adjoint functor. Here is what I know about this so far: There is a forgetful functor $U: \mathbf{Ab}\to\mathbf{CMon}$ from the category of abelian groups to the category of commutative monoids, which has a "free" left adjoint $Gr.$ If I understood correctly we can write a Kan extension formula for it as $\require{AMScd}$ \begin{CD} \mathbf{Ab} @>\text{Id}>> \mathbf{Ab}\\ @V U VV @|\\ \mathbf{CMon} @>>\text{Lan}_U(\text{Id})> \mathbf{Ab} \end{CD} By the pointwise formula $Gr(M)=\text{Lan}_U(\text{Id})(M)=\underset{\mathcal{D}}{colim}\,\text{Id}(G),$ where $\mathcal{D}$ is the diagram whose objects are $G\in\mathbf{Ab}$ with a commutative monoid homomorphism $U(G)\to M,$ and morphisms are commutative triangles. Until this point we did not use any structure of these categories. Now I wonder,

1. how to translate this into the equivalence relation that we know? and
2. what changes as we go to non-abelian groups?

Thank you in advance for your help.

Answer 1

The forgetful functor from the category of groups to the category of monoids has a left adjoint, which may be considered a non-commutative generalisation of the Grothendieck group construction. The colimit formula is not very useful for obtaining an explicit description – as you observed, it doesn't use anything special about the categories or functors in question. Here is a concrete construction.

Let $M$ be a monoid, written multiplicatively with unit $1$. We consider the group $G$ defined as follows. The elements of $G$ are finite words of the form $a b^{-1} c d^{-1} \cdots$ where $a, b, c, d, \ldots$ are elements of $M$, subject to the following equivalence relations:

• $1^{-1} a \cdots = a \cdots$ for all $a \in M$.
• $1 b^{-1} \cdots = b^{-1} \cdots$ for all $b \in M$.
• $\cdots c 1^{-1} = \cdots c$ for all $c \in M$.
• $\cdots d^{-1} 1 = \cdots d$ for all $d \in M$.
• $\cdots a 1^{-1} c \cdots = \cdots (a c) \cdots$ for all elements $a$ and $c$ in $M$.
• $\cdots a^{-1} 1 c^{-1} \cdots = \cdots (c a)^{-1} \cdots$ for all elements $a$ and $c$ in $M$.
• $\cdots (a b) (c b)^{-1} \cdots = \cdots a c^{-1} \cdots$ for all elements $a, b, c$ in $M$.
• $\cdots (a b)^{-1} (a c) \cdots = \cdots b^{-1} c \cdots$ for all elements $a, b, c$ in $M$.

The group operation of $G$ is essentially concatenation: given words $a b^{-1} \cdots$ and $c d^{-1} \cdots$, the product is $a b^{-1} \cdots c d^{-1} \cdots$, with a $1$ or $1^{-1}$ inserted in the middle if necessary. The inverse of $a b^{-1} c d^{-1} \cdots$ is $\cdots d c^{-1} b a^{-1}$. It is straightforward, if tedious, to verify that this is indeed a group such that every monoid homomorphism $M \to H$ factors as the obvious map $M \to G$ followed by a group homomorphism $G \to H$, and that this factorisation is unique.