I suppose this is a question with a well known answer. Suppose $A$ and $B$ are two algebras over some field and there is a map $$ f: \operatorname{K_0}(A) \to \operatorname{K_0}(B) $$ is it necessarily induced by a tensor product with some bimodule? If not in general, is it true for some reasonable $A$, $B$ and $f$?
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I'd say no to both questions. 1) If $k$ is a field then $K_0(k)= \mathbb Z$ generated by $[k]$ and the class of any $k$-module is positive, so $-n\colon K_0(k)\rightarrow K_0(k)$ cannot be induced by a bimodule, $n>0$. 2) If $k$ is a field of positive characteristic and $k'$ is a field of characteristic $0$ then the only map $K_0(k)\rightarrow K_0(k')$ induced by a bimodule is the trivial map, since the only $k$-$k'$-bimodule is the trivial one. The same the other way round. EDIT: Answering Sasha's comment below. If $A=\mathbb{C}[\epsilon]/(\epsilon^2)$, any left and right projective $k$-$A$-bimodule is even-dimensional over $\mathbb C$, hence all induced homomorphisms $K_0(A)=\mathbb Z\rightarrow K_0(\mathbb C) =\mathbb Z$ are multiples of $2$, in particular the identity fails to be induced. |
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In the world of $C^*$-algebras, there are cases, where a homomorphism $f \colon K_0(A) \to K_0(B)$ is induced not only by a bimodule, but by an honest $*$-homomorphism. For example, if $A$ is a separable, unital, nuclear simple In particular, if |
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