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 $f$? 



Morphisms between $K_0$
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.


