Complex topological K⁰ by Grothendieck completion

Let $X$ be compact Hausdorff and put $M={\mathrm{Vect}}_{\mathbb{C}}(X)$ as in The Whitney-sum monoid of complex vector bundles. Define

$K^0(X)=(M\times M)\mathrm{/}\sim ,$

where $(E,F)\sim (E^{\mathrm{\prime }},F^{\mathrm{\prime }})$ when there is a finite-rank complex bundle $H$ with

$E\oplus F^{\mathrm{\prime }}\oplus H\cong E^{\mathrm{\prime }}\oplus F\oplus H\mathrm{.}$

Write the class of $(E,F)$ as $[E]-[F]$. Addition and inverse are

$([E]-[F])+([E^{\mathrm{\prime }}]-[F^{\mathrm{\prime }}])=[E\oplus E^{\mathrm{\prime }}]-[F\oplus F^{\mathrm{\prime }}],-([E]-[F])=[F]-[E]\mathrm{.}$

This is the Grothendieck group of $M$. The canonical monoid map $M\to K^0(X)$ sends $[E]$ to $[E]-[0_X]$. It is universal: for every monoid map $u:M\to A$ to an abelian group there is a unique homomorphism $\bar{u}:K^0(X)\to A$ with

$\bar{u}([E]-[F])=u(E)-u(F)\mathrm{.}$

The common-summand relation is exactly what makes this displayed formula independent of the representative.