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Feb 18, 2021 at 18:15 answer added Leonid Positselski timeline score: 1
Feb 18, 2021 at 17:40 answer added Leonid Positselski timeline score: 2
Feb 18, 2021 at 12:14 answer added Leonid Positselski timeline score: 3
Feb 18, 2021 at 8:25 comment added Sam By finitely generated contramodule, I meant a finitely generated object in the category of contramodules. In this sense, can a contramodule be expressed as direct limit of its finitely generated subcontramodules? We know that a comodule over a coalgebra over a field can be expressed as direct limit of its finitely generated subcomodules (by finitely generated comodule, I mean finitely generated as a vector space which in this case is equivalent to the notion of a finitely generated object in the category of comodules). So, I am thinking if something like this holds for contramodules too.
Feb 18, 2021 at 6:50 comment added Sam I understand that all colimits exist in the category of contramodules and that finite colimits agree with those in vector spaces. I was hoping if there is a way to explicitly understand those colimits (say, arbitrary coproducts, as you mentioned) or if there is some functor that one applies on the colimit taken in the vector space to obtain the colimit in the contramodule category. For instance, the category of quasi-coherent sheaves have all arbitrary products and those are obtained by applying a functor called the 'coherator' to the product taken in larger category of sheaves of modules.
Feb 18, 2021 at 6:42 history edited Sam CC BY-SA 4.0
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Feb 16, 2021 at 19:28 comment added Leonid Positselski Finally, what do you mean by a "finitely generated contramodule"? One can say that a contramodule $P$ is finitely generated if there is a finite set of elements $p_1,\dotsc,p_n\in P$ such that every subcontramodule of $P$ containing $p_1,\dotsc,p_n$ coincides with $P$. One can also refer to the category-theoretic definition of a finitely generated object: an object $Q$ in a category $K$ is said to be finitely generated if the functor $\operatorname{Hom}_K(Q,{-})$ preserves the direct limits of diagrams of monomorphisms in $K$.
Feb 16, 2021 at 18:42 comment added Leonid Positselski Further, the colimit of any diagram in any category can be presented as the coequalizer of a suitable morphism of coproducts. So colimits can be expressed in terms of coproducts and coequalizers; and any category which has coproducts and coequalizers has all colimits. For additive categories, this reduces to the need to have coproducts and cokernels. In contramodules, the cokernels agree with those in modules or in vector spaces; the coproducts don't.
Feb 16, 2021 at 18:38 comment added Leonid Positselski Concerning the colimits, one does not "define" colimits in a specific category, like comodules or contramodules or whatever. There is a general concept of the colimit of a diagram in a category, it is defined in purely category-theoretic terms for all categories at once. The definition applies to any category, and no additional structure on a category is needed. Generally speaking, the colimits in a category may or may not exist, though. So you may want to prove that the colimits of all diagrams exist in a particular category. You may also want to know how to compute those colimits.
Feb 16, 2021 at 18:33 comment added Leonid Positselski Your definition of a contramodule is incomplete. The equation $\pi_M\circ\operatorname{Hom}(\varepsilon_C,M)=id_M$ which you wrote is the contraunitality. The contraassociativity also needs to be imposed. (In fact, if you don't impose the contraassociativity axiom, then you are not using the comultiplication in $C$ at all in your definition of a contramodule.)
Feb 16, 2021 at 14:56 history edited LSpice CC BY-SA 4.0
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Feb 16, 2021 at 12:04 history asked Sam CC BY-SA 4.0