If $A$ is a nuclear $C^*$ algebra,$A^{op}$ is the opposite $C^*$ algebra,Is the conclusion: "there is a unique injecitive $*$ homomphism $\pi:A\otimes A^{op}\rightarrow M(A)\otimes M(A)^{op}$." correct?$M(A)$ is the multiplier algebra of $A$. Login To add answer/comment