$\DeclareMathOperator\coev{coev}$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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3$\begingroup$ I don't think you want to write $r\text{coev}(1)=r\text{coev}(1)$, because this is an identity... Also, I believe $\text{coev}$ is determined by its value on $1$, by the property that it is a $R$-bimodule homomorphism, so $1\mapsto \sum e_i\otimes e^i$ extends linearly in a unique way. And once a basis of $P$ has been fixed I believe the zig-zag identities fix the value of coev(1) uniquely (I'm thinking as if $P$ was a vector space). $\endgroup$– foscoMar 19, 2022 at 20:23
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$\begingroup$ @Fosco: I have corrected the typo. Thanks for pointing it out. $\endgroup$– Adam BondalMar 19, 2022 at 20:34
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$\begingroup$ Incidentally, I think a left dual goes on the left side of the tensor in the evaluation map. $\endgroup$– Elizabeth HenningMar 21, 2022 at 16:20
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$\begingroup$ @EH: Yes of course, thank you for pointing this out. $\endgroup$– Adam BondalMar 21, 2022 at 19:20
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1$\begingroup$ @LSpice: the typo has been corrected $\endgroup$– Adam BondalMar 22, 2022 at 0:43
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