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.

  • $\begingroup$ There is no need to go into Kan extensions here. There is a forgetful functor from the category of groups to the category of monoids. It has a left adjoint. This is the non-commutative generalisation. $\endgroup$
    – Zhen Lin
    Feb 20, 2021 at 0:40
  • $\begingroup$ @ZhenLin: What would be the explicit construction of it? I like to understand how Kan extensions explain these explicit constructions. It might be helpful for me to (re)construct unknown adjoint functors in future as well. $\endgroup$
    – Bumblebee
    Feb 20, 2021 at 0:44
  • 1
    $\begingroup$ If your goal is to find an explicit construction then abstract nonsense is not going to help you. At any rate the left adjoint is easy to construct explicitly. It consists of finite "words" of the form $a b^{-1} c d^{-1} \cdots$ where $a, b, c, d, \ldots$ are elements of the monoid you start with and subject to various equivalence relations. The group operation is essentially concatenation. $\endgroup$
    – Zhen Lin
    Feb 20, 2021 at 0:52
  • $\begingroup$ @ZhenLin: Thank you for the explanation. So, these colimit formulas doesn't leads to any explicit construction? If not, what is the use of it? When would they give an explicit construction? I thought people are actually using them to understand adjoint functors. If not, how do they understand these construction in general? $\endgroup$
    – Bumblebee
    Feb 20, 2021 at 0:55
  • $\begingroup$ They are useful for proving general facts about adjoint functors. They are not usually helpful for understanding specific adjoint functors. $\endgroup$
    – Zhen Lin
    Feb 20, 2021 at 1:04

1 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.

  • $\begingroup$ Thank you very much. Is there any way that I can the constructions of these sort of left adjoint functors in general? Because most (in my limited experience) forgetful functors possess a left adjoint. $\endgroup$
    – Bumblebee
    Feb 20, 2021 at 2:46
  • $\begingroup$ There are fairly general ways of constructing left adjoints to forgetful functors, but you don’t often get concrete constructions by abstract nonsense. Call it conservation of work, if you like. $\endgroup$
    – Zhen Lin
    Feb 20, 2021 at 2:50

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