Studying K-Theory and the construction of the group $K(M)$ there are some details that still remain unclear for me (I'm reading Hatcher's book and Karoubi's book). First some background of the construction I'm using.
Let $(M, \cdot)$ be a commutative monoid. We define the abelian group $K(M)$ as the set of equivalence of classes $$K(M):= M\times M\Big/\sim \;\;,$$ where $[(m,n)]\sim[(m',n')]$ iff there exists $c\in M$ such that $m\cdot n' \cdot c = m' \cdot n \cdot c$, with $m,n \in M$.
Karoubi's book (pag. 52) says "the quotient monoid is a group". However, I could not prove that $K(M)$ is a commutative group. Especially I do not see from where the existence of inverses come from? In my opinion, this could arise from the universal property:
For any abelian group $G$, and any monoid homeomorphism $f: M\to G$, there is a unique group homeomorphism $\hat{f}: K(M) \to G$ such that the following diagram commutes.
Since $G$ is an abelian group, and a homeomorphism maps inverses to inverses, then this could imply the existence of inverses in $K(M)$.
Questions:
- How can I show clearly that $K(M)$ is an abelian group? (associativity, identity, commutativity, and especially inverses)
- If this has a relation with the universal property, how could I know that such an abelian $G$ always exists such that it is homeomorph to $M$? And further, that this construction satisfies this universal property?
Note: I know that this construction $K(X)$ is used normally for the space of equivalence classes of vector bundles over a compact Hausdorff base space $X$. But let's keep it simple and general considering just $M$ to be any commutative monoid.
Many thanks for any help!