3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.