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 commutative diagram with $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ (for $C^*$-algebras) 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 $A\in N$, $B$ 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 $Hom_{\mathbb{Z}}(P_*,K_*(B))$.

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'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). .

Edit (very simple example): If you take $A=\mathbb{C}$ and $B=K(H)$ for a separable finite dimensional complex Hilbert space $H$, it is $Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))=0=Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))$ and $KK_*(A,B)\cong Hom(K_*(A), K_*(B))\cong K^*(A)\otimes K_*(B)\cong \mathbb{Z} $.

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    $\begingroup$ You forgot the bootstrap class assumption in the Künneth theorem. $\endgroup$
    – Rasmus
    Aug 31, 2016 at 8:13

1 Answer 1


It is absolutely true that these two theorems are conceptually related. But instead of one being a consequence of the other, both rather follow from the same machinery being applied to slightly different setups: left deriving the covariant hom functor vs right deriving the contravariant hom functor. This is clarified in Example 4.5 in http://arxiv.org/pdf/math/0702146.pdf

  • $\begingroup$ thank you. I'm asking because recently I someone claimed that Künneth is formally weaker than the UCT and I want to understand it. $\endgroup$
    – user62639
    Sep 1, 2016 at 10:54

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