$\DeclareMathOperator\coev{coev}$Let $R$ be a noncommutative ring and let $P$ be a bimodule over $R$, that is finitely generated and projective as a left module. It is "well-known" that any such $P$ admits a left dual ${}^{\vee}P$ in the monoidal category of $R$-bimodules. The evaluation map $$ \operatorname{ev}:{}^{\vee}P \otimes P \to R $$ is clearly an $R$-bimodule map. The dual coevaluation map $\coev$ is given by $$ \coev: R \to P \otimes_R {}^{\vee}P, ~~ r \mapsto r \sum_{i=1}^n e_i \otimes e^i, $$ where $e_i$ and $e^i$ are dual bases for $P$ and ${}^{\vee}P$. If this is a bimodule map, then the element $\coev(1)$ would have to satisfy \begin{equation} \label{1}\tag{1} r\coev(1) = \coev(1)r, ~~ \textrm{ for all } r \in R. \end{equation} However, I cannot see that \eqref{1} is true. Also I can't see why the element $\coev(1)$ is independent of the choice of dual basis.

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