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 left $R$-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 ?()

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 ?