Category ring

Modules over a category ring

Let be a category with finite . Then a module over the the category ring is equivalent to a functor , #m/thm/rep and we have an equivalence of categories1

Proof

Let be a functor, Let and thus construct the -graded vector space

and for a morphism and homogenous vector define

and extend this definition linearly first for nonhomogenous vectors and then general . Clearly this defines a -module.

Now suppose is a natural transformation with components . Then

defines an -graded linear map. Moreover, by naturality of , for a morphism and homogenous vector

so by linearity is a -module isomorphism. Therefore is a functor.

Conversely, suppose is a -module. We define a functor as follows:

  • for ;
  • for and .

Now suppose is a -module homomorphism. We define a transformation with components

which is well-defined since is -graded. Moreover, for any and

so the following diagram commutes

https://q.uiver.app/#q=WzAsOCxbMiwyLCIoTVYpWCJdLFs0LDIsIihNVylYIl0sWzIsNCwiKE1WKVkiXSxbNCw0LCIoTVcpWSJdLFswLDIsIlgiXSxbMCw0LCJZIl0sWzIsMCwiViJdLFs0LDAsIlciXSxbMCwyLCIoTVYpZiIsMl0sWzEsMywiKE1XKWYiXSxbMCwxLCIoTVxcdmFycGhpKV9YIl0sWzIsMywiKE1cXHZhcnBoaSlfWSIsMl0sWzQsNSwiZiIsMl0sWzYsNywiXFx2YXJwaGkiXV0=

whence is natural and is a functor.

It is not difficult to see the natural equivalences required to make this an equivalence.


#state/tidy | #lang/en | #SemBr

Footnotes

  1. assuming the Axiom of Choice.