Baer's criterion for functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:40:51Z http://mathoverflow.net/feeds/question/22339 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22339/baers-criterion-for-functors Baer's criterion for functors Martin Brandenburg 2010-04-23T12:31:36Z 2010-04-24T12:03:56Z <p>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 <a href="http://maddin.110mb.com/baer.pdf" rel="nofollow">simple</a>.</p> <ul> <li>Does anyone has come across this theorem?</li> <li>Are there applications to non-representable functors?</li> <li>Can we somehow put the proof into a general pattern of the form: A statement about monomorphisms can be proven on a "generating system"?</li> <li>There are striking similarities with the proof that the cohomology of flabby sheaves vanishes. Is there a common generalization?</li> </ul> http://mathoverflow.net/questions/22339/baers-criterion-for-functors/22421#22421 Answer by unknown (google) for Baer's criterion for functors unknown (google) 2010-04-24T12:03:56Z 2010-04-24T12:03:56Z <p>Regarding the original question (with $A=R$-$\mathbf{Mod}$), I think that by SAFT <em>any</em> continuous functor $A^{\mathrm{op}}\to \mathbf{Set}$ is representable, and hence the assertion in the original question does not generalize Baer's theorem.</p> <p>In detail (with $A=R$-$\mathbf{Mod}$):</p> <p>(*) $R$ is a generator in $A$, and hence a cogenerator in $A^{\mathrm{op}}$.</p> <p>(*) $A$ 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.</p> <p>(*) $A$ is small cocomplete (as is any $\tau$-algebra, for $\tau=$(operations, identities)), and hence $A^{\mathrm{op}}$ is small complete.</p> <p>(*) Both $A^{\mathrm{op}}$ and $\mathbf{Set}$ have small hom-sets.</p> <p>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 $G$). </p> <p>(See also the discussion on Watt's theorem on p. 131 of Mac Lane).</p> <p>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 <a href="http://en.wikipedia.org/wiki/Abelian_category" rel="nofollow">Wikipedia</a> ), but I do not know enough to say anything about the question of being co-well-powered.</p>