# A Question About the Elliott-Natsume-Nest Proof of Bott Periodicity

In Wegge-Olsen’s book K-Theory and C$^{*}$-Algebras, there is an outline of a proof of Bott Periodicity (the proof is due to George Elliott, Toshikazu Natsume and Ryszard Nest). The first step of the proof outline goes as follows:

Suppose that there are natural transformations

• $\Phi^{0}$ from the $K_{0} S$-functor to the $K_{1}$-functor and
• $\Phi^{1}$ from the $K_{1} S$-functor to the $K_{0}$-functor,

where $S$ denotes the suspension functor. In other words, for every C$^{*}$-algebra $\mathscr{A}$ (not necessarily unital), there are abelian-group homomorphisms $$\Phi^{0}_{\mathscr{A}}: {K_{0}}(S(\mathscr{A})) \to {K_{1}}(\mathscr{A}) \quad \text{and} \quad \Phi^{1}_{\mathscr{A}}: {K_{1}}(S(\mathscr{A})) \to {K_{0}}(\mathscr{A})$$ such that for every C$^{*}$-algebraic homomorphism $\alpha: \mathscr{A} \to \mathscr{B}$, the following diagrams commute: $$\require{AMScd} \begin{CD} {K_{0}}(S(\mathscr{A})) @>{{K_{0} S}(\alpha)}>> {K_{0}}(S(\mathscr{B})) \\ @V{\Phi^{0}_{\mathscr{A}}}VV @VV{\Phi^{0}_{\mathscr{B}}}V \\ {K_{1}}(\mathscr{A}) @>>{{K_{1}}(\alpha)}> {K_{1}}(\mathscr{B}) \end{CD} \quad \quad \quad \begin{CD} {K_{1}}(S(\mathscr{A})) @>{{K_{1} S}(\alpha)}>> {K_{1}}(S(\mathscr{B})) \\ @V{\Phi^{1}_{\mathscr{A}}}VV @VV{\Phi^{1}_{\mathscr{B}}}V \\ {K_{0}}(\mathscr{A}) @>>{{K_{0}}(\alpha)}> {K_{0}}(\mathscr{B}) \end{CD}$$ Suppose further that $\Phi^{0}_{\mathbb{C}}$ and $\Phi^{1}_{\mathbb{C}}$ are isomorphisms. Then prove that $\Phi^{0}_{\mathscr{A}}$ and $\Phi^{1}_{\mathscr{A}}$ are isomorphisms for every C$^{*}$-algebra $\mathscr{A}$.

Wegge-Olsen’s hint is to consider the commutative diagrams above for the homomorphism \begin{align*} \alpha: \mathbb{C} & \to \mathscr{A} \otimes \mathbb{K}; \\ 1 & \mapsto p \otimes e_{11}, \end{align*} where $p$ is a fixed choice of a projection in $\mathscr{A}$ and $e_{11}$ is a rank-$1$ projection in $\mathbb{K} \stackrel{\text{def}}{=} {M_{\infty}}(\mathbb{C})$.

My difficulty:

I do not see how \begin{align*} {K_{0}}(\alpha): & {K_{0}}(\mathbb{C}) \to {K_{0}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{1}}(\alpha): & {K_{1}}(\mathbb{C}) \to {K_{1}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{0} S}(\alpha): & {K_{0}}(S(\mathbb{C})) \to {K_{0}}(S(\mathscr{A} \otimes \mathbb{K})) \quad \text{and} \\ {K_{1} S}(\alpha): & {K_{1}}(S(\mathbb{C})) \to {K_{1}}(S(\mathscr{A} \otimes \mathbb{K})) \end{align*} are isomorphisms. If this hurdle can be overcome, then $\Phi^{0}_{\mathscr{A}}$ and $\Phi^{1}_{\mathscr{A}}$ are clearly isomorphisms.

Thank you for any assistance!

• Obviously, $K_0(\alpha)$ and the rest of the mappings are not isomorphisms. Because otherwise it would imply that all $C^\ast$-algebras have isomorphic $K_0$-groups (and $K_1$-groups). My guess is that he has shown that the suspension functor is surjective. Then using the above technique he tries to show that the suspension is injective as well. – user23860 Jan 29 '14 at 11:45

Consider something by putting your two Diagramms together (for the map $\alpha:\mathbf{C} \rightarrow A$) so that they fuse to one diagram with three lines. In the first diagram you replace $A$ by $S(A)$ so that double Suspension $SS(A)$ is involved.
The first column is then an isomorphism $K_0(SS(\mathbf{C}) \cong K_0(\mathbf{C})$.
Chase in the Diagramm, to see that the second column is also something (a isomorphism (??!)) for the subgroup generated by $[p]$. Then vary over all $p$.
• I was thinking about this problem too. Following the suggestion in this answer, I think I can get the surjectivity of $\Phi^0_A$ and $\Phi^1_A$ for all $A$, but I couldn't figure out the injectivity. – cyc Jan 15 '18 at 22:26