# Question on Cuntz' proof of Bott periodicity

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?

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.