The Grothendieck group construction

In Lang, the example is stated as follows:

Let M be a commutative monoid, written additively. There exists a commutative group K(M) and a monoid homomorphism.

$$\gamma: M\rightarrow K(M)$$

having the following universal property if $f:M\rightarrow A$ is a homomorphism into an abelian group A, then there exists a unique homomorphism $f_*:K(M)\rightarrow A$ making the following diagram commutative, $f=f_*\circ \gamma$

Proof: Let $F_{ab}(M)$ be the free abelian gropu generated by M. We denote the generator of $F_{ab}(M)$ corresponding to an element $x\in M$ by $[x]$. Let B be the subgroup generated by all elements of type $$[x+y]-[x]-[y]$$

where $x,y\in M$. We let $K(M)=F_{ab}(M)/B$, and let $$\gamma:M\rightarrow K(M)$$

be the map obtained by composing the injection of $M$ into $F_{ab}(M)$ given $x\mapsto [x]$, and the canonical map $$F_{ab}(M)\rightarrow F_{ab}(M)/B$$

It is then clear that $\gamma$ is a homomorphism, and satisfies the desired universal property.

Question: What I don't understand is why we factor out the group generated by all elements of the type $[x+y]-[x]-[y]$. Is it similar to how the kernel of all homorphisms from a group into an abelian group factors through the commutator of the first group?

• It's similar but a bit more basic. You simply want the relation $\gamma(x+y)=[x+y]=[x]+[y]=\gamma(x)+\gamma(y)$ to hold in $K(M)$ or otherwise the map $\gamma: M \rightarrow K(M)\ x\mapsto [x]$ will not be a homomorphism. The point is that's exactly the minimum such requirement. Any other abelian group $A$ with a map $\eta: M \rightarrow A$ will also satisfy $\eta(x+y)=\eta(x)+\eta(y)$ so the map $[x] \mapsto \eta(x)$ from $F_\textit{ab}(M)$ to $A$ will also be well-defined and $\eta$ will factor through it. – Zavosh Sep 15 '14 at 1:05
• That's an answer. – Kevin Carlson Sep 15 '14 at 1:10
• Alright thanks! – Enigma Sep 15 '14 at 2:48