I'm crrently reading the paper about the Künneth-theorem for $C^*$-algebras: http://msp.org/pjm/1982/98-2/pjm-v98-n2-p15-s.pdf and I'm trying to understand remark 4.9. I henceforth asumme that $A$ is a separable nuclear $C^*$algebra and $B$ is a separable $C^*$algebra. Furthermore, I will adopt all notations from the paper.
We assume that:
(1). If $K_*(B)$ is free and if $f:A'\to A''$ is a $*$-homomorphism inducing an isomorphism $f_*:K_*(A')\to K_*(A'')$, then $(f\otimes 1)_*:K_*(A'\otimes B)\to K_*(A''\otimes B)$ is an isomorphism.
And we want to show that:
(2). Künnth-theorem (theorem 4.1) holds in full generality.
(As mentioned in the paper, it suffices to assume that $A$,$B$ are separable $C^*$ algebras with $A$ nuclear and $K_*(B)$ torsion free and to show that $\alpha(A,B)$ is an isomorphism).
What I have done so far: Assume (1) and that $K_*(A)$ is free. Then I obtain that (2) holds with $K_*(A \otimes B) \cong K_*(A) \otimes K_*(F)$. However, I the assumption (that $K_*(A)$ is free) is too strong and the result is somehow strange. My try was the following: I tried to prove $\operatorname{coker}(1 \otimes v_*)=\operatorname{coker} ((1 \otimes v)_*)$ and $\ker(1 \otimes v_*)=\ker ((1 \otimes v)_*)$ as in proof of thm 4.1 and for this I'm using diagram (4.5). But I used that $\alpha(A,F)$ is an isomorphism (which is correct even though if we would drop that $A$ is in the bootstrap-class) to use that $K_*(A) \otimes K_*(F) \cong K_*(A \otimes F)$.
First I considered the top-row in diagram (4.5) in the proof of theorem 4.1: Since $K_*(B)$ is free it is $\operatorname{Tor}(K_*(A),K_*(B))=0$ and $\mu_*:K_*(F)\to K_*(B)$ is an isomorphism, it follows that $1\otimes ( \mu)_*:K_*(A) \otimes K_*(F)\to K_*(A) \otimes K_*(B)$ is an isomorphism as well. Thus, $K_*(A)\otimes K_*(C)=0$ (since the sequence is exact) and $1 \otimes v_*=0$.
Now, if we assume that $A$ nuclear and that $K_*(A)$ is free, then we can apply (1) to $\mu$ as follows :
since $\mu_*:K_*(F)\to K_*(B)$ is an isomorphism, $(\mu \otimes 1_A)_*:K_*(F \otimes A)\to K_*( B \otimes A)$ is an isomorphism.
Applying the 6-term exact sequence, it follows that $K_*(C \otimes A) =0$.
And then we have in the top row of this diagram that $(1 \otimes v)_*=0$.
It follows that $\ker(1 \otimes v_*)=\ker ((1 \otimes v)_*) =0$ and:
$\operatorname{coker}(1 \otimes v_*) \cong [K_*(A) \otimes K_*(F)]/Im (1 \otimes v_*)=K_*(A) \otimes K_*(F)$,
$\operatorname{coker} ((1 \otimes v)_*) \cong K_*(A \otimes F)/\operatorname{Im}((1 \otimes v)_*)\cong K_*(A \otimes F)$, hence
$\operatorname{coker}(1 \otimes v_*)=\operatorname{coker} ((1 \otimes v)_*)$.
Thank you for any help.
