The complex K-ring of S²

By [F1], $K^0(S^2)$ is generated by $1$ and $\beta$, the only polynomial relation is ${\beta }^2=0$, and both generators are additively independent. This proves both displayed descriptions, including the zero reduced summand only when its integer coefficient is zero.

The class of $\gamma$ is $1+\beta$. Since ${\gamma }^{\ast }$ is the tensor inverse of $\gamma$, its class is the multiplicative inverse of $1+\beta$. The relation ${\beta }^2=0$ gives $(1+\beta )(1-\beta )=1-{\beta }^2=1$, so $[{\gamma }^{\ast }]=1-\beta$ and $[{\gamma }^{\ast }]-1=-\beta$.

By [F2], reversing the hemispheres changes the loop $z$ to $z^{-1}$, which clutches the dual line ${\gamma }^{\ast }$. Step 2.1 therefore proves that this convention reverses the Bott generator and leaves the square-zero presentation unchanged.