I have a (maybe dumb) question about the relation between the Künneth theorem and the Theorem-universal coefficient theorem (UCT for short) in $KK$-theory (for the setting see "The Künneth theorem and the universal coefficient theorem for Kasparov’s generalized K-functor"- Rosenberg and Schochet). 
 
Let $N$ be the bootstrap class of $C^*$-algebras (i.e. see here http://mathoverflow.net/questions/241884/commutative-diagram-with-k-i1a-to-k-ia-rtimes-rho-mathbbr-for-c  for the definition). 

**UCT**: Let $A\in N$ and $B$ be $\sigma$-unital. Then there is a short exact sequence $$0\to Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))\to KK_*(A,B)\to Hom(K_*(A), K_*(B))\to 0.$$

**Künneth**: Let $A$, $B$ be $\sigma$-unital and $K_*(B)$ finitely generated. Then there is a short exact sequence $$0\to K^*(A)\otimes K_*(B)\to KK_*(A,B)\to Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))\to 0.$$


>>Is it possible to proof Künneth $\Rightarrow $ UCT, and if so, how to prove it (do you have a reference)? What is already known?

Edit: In the paper the theorems are proved independently. 

By definition it is $Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))=H^1(Hom_{\mathbb{Z}}(P_*,K_*(B)) [\cong H^1(Hom_{\mathbb{Z}}(K_*(A),I^*))\enspace ]$, where $P_*$ is a projective resolution of $K_*(A)$ and $I^*$ an injective resolution of $K_*(B)$. $H^1$ denotes the first cohomology of the cochain complex $Hom_{\mathbb{Z}}(P_*,K_*(B))$.
Thus, is $H^1(Hom_{\mathbb{Z}}(P_*,K_*(B))\cong K^*(A)\otimes K_*(B)$ for some $A$ and $B$?. 


Then it is $Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))=H_1(P_*\otimes K_*(B)) [\cong H_1(K^*(A)\otimes D_*))]$, where $P_*$ is a projective resolution of $K^*(A)$ and $D_*$ is a projective resolution of $K_*(B)$. $H_1$ is the first homology of the chain complex $P_*\otimes K_*(B)$. I.e. is $H_1(P_*\otimes K_*(B))\cong Hom(K_*(A), K_*(B))$ in some situations?

I'm not sure if in this case we can use that homology and cohomology commute with exact functors (since the tensor product is exact for $P_*$ projective and Hom is exact for $P_*$ projective and $I^*$ injective respectively). .