Modules over a category ring
Let
Proof
Let
be a functor, Let πΉ : π’ β π΅ πΎ πΌ π π and thus construct the ( π πΉ ) π = πΉ π -graded vector space O b β‘ ( π’ ) π πΉ = β¨ π β O b β‘ ( π’ ) ( π πΉ ) π and for a morphism
and homogenous vector π β π’ define π£ β ( π πΉ ) π π β π£ = { ( πΉ π ) π£ π = d o m β‘ π 0 π β d o m β‘ π and extend this definition linearly first for nonhomogenous vectors and then general
. Clearly this defines a π β π [ π’ ] -module. π [ π’ ] Now suppose
is a natural transformation with components π β π΅ πΎ πΌ π π π’ ( πΉ , πΊ ) . Then π π : ( π πΉ ) π β ( π πΊ ) π ( π π ) = β¨ π β O b β‘ ( π’ ) π π : π πΉ β π πΊ defines an
-graded linear map. Moreover, by naturality of O b β‘ ( π’ ) , for a morphism π and homogenous vector π β π’ ( π , π ) π£ β ( π πΉ ) π ( π π ) ( π β π£ ) = π π ( πΉ π ) π£ = ( πΊ π ) π π π£ = π β ( π π ) π£ so by linearity
is a π π -module isomorphism. Therefore π [ π’ ] is a functor. π : π΅ πΎ πΌ π π π’ β π [ π’ ] π¬ π π½ Conversely, suppose
is a π -module. We define a functor π [ π’ ] as follows: π π : π’ β π΅ πΎ πΌ π π
for ( π π ) π = 1 π β π ; π β O b β‘ ( π’ ) for ( ( π π ) π ) π£ = π β π£ and π β π’ ( π , π ) . π£ β ( π π ) π Now suppose
is a π : π β π -module homomorphism. We define a transformation with components π [ π’ ] ( π π ) π : ( π π ) π β ( π π ) π π£ β¦ π π£ which is well-defined since
is π -graded. Moreover, for any O b β‘ ( π’ ) and π β π’ ( π , π ) π£ β π ( π ) π ( ( π π ) π ) ( π π ) π π£ = π β π π£ = π ( π β π£ ) = ( π π ) π ( ( π π ) π ) so the following diagram commutes
whence
is natural and π π is a functor. π : π [ π’ ] π¬ π π½ β π΅ πΎ πΌ π π π’ It is not difficult to see the natural equivalences required to make this an equivalence.
Footnotes
-
assuming the Axiom of Choice. β©