# Baer's criterion for functors

Baer's criterion can be generalized as follows: Let $A$ be an abelian category satisfying (AB3-5) with a generator $R$ and let $T : A^{o} \to Set$ be a continuous functor such that $T(R) \to T(I)$ is surjective for all subobjects $I \subseteq R$. Then for every monomorphism $M \to N$ the map $T(N) \to T(M)$ is surjective. If $T$ is representable and $A=R-Mod$, this becomes the usual Baer's criterion. The proof is simple.

• Does anyone has come across this theorem?
• Are there applications to non-representable functors?
• Can we somehow put the proof into a general pattern of the form: A statement about monomorphisms can be proven on a "generating system"?
• There are striking similarities with the proof that the cohomology of flabby sheaves vanishes. Is there a common generalization?
• I have some questions: 1. Can you link the proof Your link is broken? 2. How doest the proof implies the Baer's criterion? – Ivan Di Liberti Jun 6 '18 at 15:59

Regarding the original question (with $A=R$-$\mathbf{Mod}$), I think that by SAFT any continuous functor $A^{\mathrm{op}}\to \mathbf{Set}$ is representable, and hence the assertion in the original question does not generalize Baer's theorem.

In detail (with $A=R$-$\mathbf{Mod}$):

(*) $R$ is a generator in $A$, and hence a cogenerator in $A^{\mathrm{op}}$.

(*) $A$ is co-well-powered, because there is a bijection between the quotient objects of $M\in A$ and the set of submodules of $M$, and the latter set is small (since by assumption $M$ is small). It follows that $A^{\mathrm{op}}$ is well-powered.

(*) $A$ is small cocomplete (as is any $\tau$-algebra, for $\tau=$(operations, identities)), and hence $A^{\mathrm{op}}$ is small complete.

(*) Both $A^{\mathrm{op}}$ and $\mathbf{Set}$ have small hom-sets.

So, all the conditions of SAFT hold for a functor $A^{\mathrm{op}}\to\mathbf{Set}$, and hence any such continuous functor has a left adjoint. Now, if a functor $G\colon A^{\mathrm{op}}\to\mathbf{Set}$ has a left adjoint then it is surely representable: Saying that a functor $G$ is representable is like saying that there is a universal arrow from a one-object set $1$ to $G$ (Prop. 3.2.2, p. 60 in Mac Lane), and for this we can take the unit $\eta_1\colon 1\to G(F1)$ (with $F$ the left adjoint of $G$).

(See also the discussion on Watt's theorem on p. 131 of Mac Lane).

I am not sure about the general case of the edited question (where $A$ is an arbitrary abelian category with a generator + AB3--AB5). Cocompleteness holds by AB3 (as I have seen in Wikipedia ), but I do not know enough to say anything about the question of being co-well-powered.

• somehow I skipped so far these basics. thank you!! – Martin Brandenburg Apr 24 '10 at 23:30