Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?
share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
somehow I skipped so far these basics. thank you!! –  Martin Brandenburg Apr 24 '10 at 23:30
    
You're welcome! –  user2734 Apr 25 '10 at 4:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.