$\eta$-completed and $\eta$-inverted $KGL$

Question

Let $\eta \in \pi_{1, 1}\mathbb{S}$ denote the motivic Hopf map and $KGL$ be the motivic ring spectrum representing algebraic K-theory. I have been trying to show that $KGL_\eta^\wedge = KGL$, which in turn will tell me that $KGL[\eta^{-1}]=0$, but I am getting stuck. I know that $\pi_{1, 1}KGL=0$, thus composing $\eta$ with the unit map of $KGL$ gives me a map $S^{1, 1} \to S^{0, 0} \to KGL$ which must be nullhomotopic. This tells me that the unit map must factor through the cofiber of $\eta$. After here I am stuck; I know that $\eta$-completion is the same as localization at $\eta$ but I don't know much else.

Answer 1

If I understand it correctly there is a fiber sequence $$\mathrm{holim}_n S^{n,n}\otimes X\to X \to X^\wedge_\eta,$$ where the limit is taken over the maps $$S^{n,n}\otimes X \simeq S^{n-1,n-1}\otimes S^{1,1}\otimes X\xrightarrow{ \mathrm{id}\otimes \eta\otimes \mathrm{id}}S^{n-1,n-1}\otimes S^{0,0}\otimes X\simeq S^{n-1,n-1}\otimes X.$$ If $X$ is a ring spectrum the equivalence $S^{0,0}\otimes X\simeq X$ is homotopic to $S^{0,0}\otimes X\to X\otimes X\to X$, the composition of the unit map with the multiplication. In the case $X=KGL$ composing this with $$S^{1,1}\otimes KGL\xrightarrow{\eta\otimes\mathrm{id}} S^{0,0}\otimes KGL$$ yields the zero map as you note, and the sequential homotopy limit over zero maps is the zero object, which shows that $KGL\simeq KGL^\wedge_\eta$.