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Let $A,B,C$ be $C^\ast$-algebras. Suppose $B$ and $C$ to be strongly morita equivalent. Then $KK(A,B)\cong KK(A,C)$.

Could someone provide a reference or proof of this fact?

I guess the imprimitivity bimodule defines an element in $KK(B,C)$ and Kasparov product with this element is the desired isomorphism, but I don't know how to fill the details. Also I am unsure whether I should add $\sigma$-unitality/separability as hypothesis.

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2 Answers 2

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In Blackadar's K-Theory for Operator Algebras (first edition), paragraph 13.7.1 is an exercises that guides you to prove that if $A$ and $B$ are strongly Morita equivalent $\sigma$-unital C*-algebras, then $A$ and $B$ are stably isomorphic. Hence use the fact that KK-Theory is stable in both arguments to get to the sought isomorphism.

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  • $\begingroup$ Thanks for the info. Nevertheless, I would like a more direct proof, like the one I was sketching in the question. $\endgroup$
    – vap
    Apr 7, 2015 at 13:40
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    $\begingroup$ @vap Mybe a bit late but i came along with the same question. Did you already find a direct proof like you was looking for? I am looking for a proof that avoids stability of KK, such that I can PROVE stability using Morita equivalence. For the Kasparov product I think it is straightforward that the dual imprimitivity bimodule is the inverse. $\endgroup$
    – Roland
    Jun 8, 2023 at 11:27
  • $\begingroup$ @Roland Please read the answer I just wrote. $\endgroup$
    – vap
    Jun 8, 2023 at 11:59
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This is proved in Theorem 2.18 of G. G. Kasparov, "Equivariant KK-theory and the Novikov conjecture", Invent. Math. 91 (1988), 147–201. The algebras $B$ and $C$ are $\sigma$-unital, and no assumption is required of $A$. By this work by R. Exel (Theorem 5.3), $B$ and $C$ have isomorphic $K$-groups (even when they are not $\sigma$-unital).

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  • $\begingroup$ Let me clarify something: Your question is already answered by the first part of Kasparov 2.18. The proof of this is straightforward and needs the Kasparov product. Therefore I wonder if the stability of KK is needed somewhere in the Kasparov product (I guess no?). If not then stability of KK can be proved by the Cuntz picture as in Ralf Meyer "Equivariant Kasparov theory and generalised homomorphisms" Thm 5.5 (here we need that Morita equivalent $C^*$-algebras are KK-equivalent). The right handside in 5.5 is stable since the Hilbert module $G\mathbb{N}\otimes A$ absorbs $G$-Hilbert spaces. $\endgroup$
    – Roland
    Jun 8, 2023 at 14:14
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    $\begingroup$ @Roland The proof doesn't really use full power of the Kasparov product because there are no operators involved, it boils down to a tensor product of bimodules. $\endgroup$
    – vap
    Jun 9, 2023 at 9:17

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