Grothendieck ring structure and rank map

Tensor product distributes over Whitney sum, so it extends through the Grothendieck completion to a commutative unital multiplication on $K^0(X)$. On virtual-bundle representatives,

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

The unit is the trivial complex line $[{\epsilon }^1]$. Together with the addition in Complex topological K⁰ by Grothendieck completion, this makes $K^0(X)$ a commutative ring.

Fiber dimension is topologically locally constant. Define the rank map

$\mathrm{rk}:K^0(X)⟶H^0(X;\mathbb{Z})$

by

$\mathrm{rk}([E]-[F])(x)={\dim }_{\mathbb{C}}{E}_x-{\dim }_{\mathbb{C}}{F}_x\mathrm{.}$

Here singular $H^0(X;\mathbb{Z})$ is identified, as in Singular cohomology ring, with integer-valued functions constant on path components. A topologically locally constant rank function is constant along every path and hence defines such a class. Direct sum and tensor product give pointwise addition and multiplication of ranks, so $\mathrm{rk}$ is a unital ring homomorphism. No assertion that path components are open is needed.