Morphisms between $K_0$ 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 $f$? 
 A: 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 $C^*$-algebra and $B$ is separable and nuclear and if the universal coefficient theorem holds for the pair $A,B$, then $f$ is the image of some element in $KK(A,B)$ with respect to the map $KK_*(A,B) \to Hom(K_*(A),K_*(B))$ (as defined for example in the book by Blackadar) and any such element comes from a full homomorphism $\varphi \colon A \otimes \mathcal{O}_{\infty} \otimes \mathbb{K} \to B \otimes \mathcal{O}_{\infty} \otimes \mathbb{K}$ by Phillips' classification theorem (Theorem 4.1.3 in that paper. Here $\mathbb{K}$ denotes the compact operators on a separable infinite dimensional Hilbert space and $\mathcal{O}_{\infty}$ is the infinite Cuntz algebra.). 
In particular, if $B \cong B \otimes \mathcal{O}_{\infty}$, $A \cong A \otimes \mathcal{O}_{\infty}$ and if you can lift $f$ to an invertible element in $KK(A,B)$, then $f$ comes from a stable isomorphism $\varphi \colon A \otimes \mathbb{K} \to B \otimes \mathbb{K}$, which in turn corresponds to a Morita equivalence between $A$ and $B$ by a theorem of Brown, Green and Rieffel. The conditions $A \cong A \otimes \mathcal{O}_{\infty}$ and $B \cong B \otimes \mathcal{O}_{\infty}$ are for example satisfied if $A$ and $B$ are simple and purely infinite $C^*$-algebras.
A: 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.
