Defining the dual-comodule of a comodule?

Question

As is well known, every left module has a dual, which is a right module. How does this work for comodules? More explicitly, does there exist a notion of the dual-comodule of a comodule? Does right and left get interchanged as in the module case.

Answer 1

Given any coalgebra $C$ over a field ${\mathbb k}$, the category of $C$-comodules fully embeds into the category of $C^{\ast}$-modules by sending a $C$-comodule $(M, \nabla: M\to C\otimes M$) to the $C^{\ast}$-module having $M$ as the underlying ${\mathbb k}$-vector space, and with $C^{\ast}$-action given by $C^{\ast}\otimes M\to C^{\ast}\otimes C\otimes M\to M$. See Brzezinski-Wisbauer, Corings and Comodules, §4.

The duality on $C^{\ast}\text{-mod}$ will in general not preserve $C$-comodules. As an example, consider $C := {\mathbb k}[t]$. Then $C^{\ast}={\mathbb k}[[t]]$, and $C\text{-coMod}$ identifies with torsion-modules over ${\mathbb k}[[t]]$, i.e. those modules $M$ for which for any $m$ there exists some $k\gg 0$ such that $t^k m = 0$, and those are not stable under duality. For example, the regular $C$-comodule $C$ is sent to the ${\mathbb k}[[t]]$-module ${\mathbb k}((t))/{\mathbb k}[[t]]$, but its ${\mathbb k}$-dual as a ${\mathbb k}[[t]]$-module is just the regular ${\mathbb k}[[t]]$-module, which has no torsion at all.

Still, generalizing the torsion-submodule functor, the embedding $C\text{-coMod}\to C^{\ast}\text{-Mod}$ has a right adjoint (see §7 in Brzezinski-Wisbauer), i.e. any $C^{\ast}$-module admits a maximal submodule coming from a $C$-comodule. Using this coreflection, you might define the restricted dual of any $C$-comodule. However, recall from the example $C=M={\mathbb k}[t]$ above that this restricted dual might even vanish.