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?



Question on Cuntz' proof of Bott periodicity
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.


