# Induced map in K-theory by a "trivial" bimodule

Let $$R$$ be a ring (not necessary commutative) and let $$P_{\bullet}$$ be a perfect $$R$$-bimodule (chain complex). I will denote the category of perfect right $$R$$-chain complexes by $$\textbf{Perf}(R)$$. The endofunctor $$-\otimes_{R}P_{\bullet} :\textbf{Perf}(R)\rightarrow \textbf{Perf}(R)$$ induces a map in algebraic $$K$$-theory given by

$$K_{\ast}(-\otimes_{R}P_{\bullet}):K_{\ast}(R)\rightarrow K_{\ast}(R)$$.

If the class $$[P_{\bullet}] \in K_{0}(R)$$ is trivial $$(=0)$$ does it mean that $$K_{\ast}(-\otimes_{R}P_{\bullet})$$ is a 0 map ?

• When you write $[P_{\bullet}] \in K_{0}(R)$, do you mean the class of $P_\bullet$ considered as a complex of right $R$-modules (forgetting the left $R$-module structure)? Jun 14 '20 at 20:09
• @JeremyRickard Yes Jun 14 '20 at 20:14

No. Let $$R=\mathbb{Z}\times\mathbb{Z}$$, let $$P$$ and $$Q$$ be the projective modules $$\mathbb{Z}\times0$$ and $$0\times\mathbb{Z}$$, and let $$P_\bullet=\dots\longrightarrow0\longrightarrow P\otimes_\mathbb{Z}P \stackrel{0}{\longrightarrow}Q\otimes_\mathbb{Z}P\longrightarrow0\longrightarrow\dots$$