Skip to main content
added 4 characters in body
Source Link
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator $P$, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$).

So if you want more general examples than this then $G$ can't satisfy the solution set condition.

I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$).

So if you want more general examples than this then $G$ can't satisfy the solution set condition.

I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator $P$, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$).

So if you want more general examples than this then $G$ can't satisfy the solution set condition.

Source Link
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$).

So if you want more general examples than this then $G$ can't satisfy the solution set condition.