Question on Cuntz' proof of Bott periodicity

Question

I am reading the presentation of Cuntz' proof of Bott periodicity for $C^*$-algebras in Wegge-Olsen (Thm. 11.2.1). Here one considers the short exact sequence of $C^*$-algebras $$0 \longrightarrow \mathcal{T}_0 \longrightarrow \mathcal{T} \stackrel{q}{\longrightarrow} \mathbb{C} \longrightarrow 0,$$ where $\mathcal{T}$ is the Toeplitz algebra and $q$ is the homomorphism determined by sending the shift operator $S \in \mathcal{T}$ to $1$. Clearly, we also have the map $j: \mathbb{C} \rightarrow \mathcal{T}$, and $q \circ j = \mathrm{id}_{\mathbb{C}}$.

Now, the main step in the proof is to show that the induced map in $K$-theory $q_*: K_0(\mathcal{T}) \rightarrow K_0(\mathbb{C})$ is an isomorphism. But then, it is claimed that it was also clear that also for any $C^*$-algebra $A$, one has that also $$(\mathrm{id}_A \otimes q)_*: K_0(A \otimes \mathcal{T}) \rightarrow K_0(A)$$ is an isomorphism.

I do understand that since all $C^*$-algebras in the short exact sequence are nuclear, the sequence $$0 \longrightarrow A \otimes \mathcal{T}_0 \longrightarrow A \otimes \mathcal{T} \stackrel{\mathrm{id}_A \otimes q}{\longrightarrow} A \longrightarrow 0$$ is exact for any $C^*$-algebra. But I do not see why knowing that $q_*$ is an isomorphism implies the same thing for $(\mathrm{id}_A \otimes q)_*$. Any suggestions?

Answer 1

The conclusion drawn in the book of Wegge-Olsen is wrong (explained below), but can, however, easily be tweaked to a correct proof. What is shown is that $j\circ q$ is homotopic to the identity on $\mathcal T$ and hence the same is true after tensoring with $A$. It follows that $K_\ast(\mathcal T_0 \otimes A) = 0$ for every $C^\ast$-algebra $A$.

A quotient map $q$ of $C^\ast$-algebras is an isomorphism in $K$-theory if and only if $K_\ast(\ker q) =0$ (by six-term exactness). So the statement that a quotient map $q$ (with splitting $j$) induces an isomorphism in $K$-theory implies that $(q\otimes_\alpha id_A)_\ast$ is an isomorphism is equivalent to the statement that $K_\ast(B)=0$ implies $K_\ast(B \otimes_\alpha A) = 0$ (for $\otimes_\alpha$ being either the maximal or minimal tensor product). And this is in general wrong. A $C^\ast$-algebra $A$ satisfies the maximal (respectively minimal) Künneth theorem if and only if $K_\ast(B \otimes_{\max} A) = 0$ (resp. $K_\ast(B \otimes_{\min{}} A) = 0$) for every $C^\ast$-algebra $B$ with $K_\ast(B) = 0$, see https://arxiv.org/abs/1111.7228 Theorems 3.1 and 4.1. This paper also contains examples of $C^\ast$-algebras that don't satisfy the maximal/minimal Künneth theorem. These examples are modelled after Skandalis' counterexamples to the Universal Coefficient Theorem.