# The coevaluation map for a projective module and its dual

$$\DeclareMathOperator\coev{coev}$$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.

• 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). Mar 19, 2022 at 20:23
• @Fosco: I have corrected the typo. Thanks for pointing it out. Mar 19, 2022 at 20:34
• Incidentally, I think a left dual goes on the left side of the tensor in the evaluation map. Mar 21, 2022 at 16:20
• @EH: Yes of course, thank you for pointing this out. Mar 21, 2022 at 19:20
• @LSpice: the typo has been corrected Mar 22, 2022 at 0:43