Let $C$ be a coalgebra over a commutative ring $R$, if $C$ is cauchy (f.g. and projective), then there is an equivalence of categories between $\operatorname{Comod}(C)$ and the category $\operatorname{Mod}(C^*)$ of modules over the dual algebra $C^*=\operatorname{hom}_R(C,R)$ (See https://mathoverflow.net/questions/94115/when-a-comodule-category-is-equivalent-to-a-module-category )

If $C$ is itself a hopf algebra then this categories are both monoidal with tensor product given by $\otimes_R$. My question: is this equivalence of categories monoidal?

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    $\begingroup$ You mean an equivalence between the category of right $C$-comodules and the category of left $C^\ast$-modules? (Or the other way round, but, importantly, not "right/right" or "left/left".) If so, I think this equivalence is actually an isomorphism which takes any right $C$-comodule $M$ and sends it to the left $C^\ast$-module $M$ whose action is given by $f m = m_{(0)} f\left(m_{(1)}\right)$ for all $f \in C^\ast$ and $m \in M$. It is fairly clear that this isomorphism preserves tensor products when $C$ is a bialgebra. $\endgroup$ – darij grinberg Jun 9 '15 at 23:04

It is an isomorphism of categories. Moreover, if C is a Bialgebra, it is symmetric monoidal. I checked it a few minutes ago: It is very easy, but only if you have DIN A3 paper.


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