Hopf-line calculation of K⁰(S²)

Let $E$ have rank $n>0$. By [F1] it is clutched by a loop $g:S^1\to {\mathrm{GL}}_n(\mathbb{C})$. If $k$ is the winding number of $\det g$, [F2] deforms $g$ to $\mathrm{diag}(z^k,1,\dots ,1)$, so [F1] gives $E\cong {\gamma }^k\oplus {\epsilon }^{n-1}$. The rank-zero bundle is $0_{S^2}$. Consequently every virtual class is an integral combination of $1$ and powers of $[\gamma ]$.

The loops $\mathrm{diag}(z^2,1)$ and $\mathrm{diag}(z,z)$ have the same determinant. By [F2] they are homotopic, and [F1] gives ${\gamma }^2\oplus {\epsilon }^1\cong \gamma \oplus \gamma$. Hence $[\gamma {]}^2-2[\gamma ]+1=0$, or ${\beta }^2=0$. It follows algebraically that $[\gamma {]}^k=(1+\beta {)}^k=1+k\beta$ for every $k\in \mathbb{Z}$, since $(1+\beta {)}^{-1}=1-\beta$.

If $a+k\beta =0$, restriction to a point gives $a=0$. Then $k\beta =0$ implies $[{\gamma }^k]=1$ by step 1.2. By [F3], after adding the same trivial bundle, ${\gamma }^k$ and the trivial line are isomorphic. Their stabilized clutching determinants have winding numbers $k$ and $0$, so [F2] forces $k=0$. Thus $1$ and $\beta$ are additively independent. Together with steps 1.1–2.1 this proves the displayed ring presentation.

Basepoint restriction sends $1$ to $1\in \mathbb{Z}$ and $\beta =[\gamma ]-1$ to $0$, so its kernel is exactly $\mathbb{Z}\beta$. Reversing the hemisphere convention replaces $z$ by $z^{-1}$ and hence $\gamma$ by ${\gamma }^{\ast }$; step 1.2 gives $[{\gamma }^{\ast }]-1=(1-\beta )-1=-\beta$.