Monomorphisms in functor categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:18:36Z http://mathoverflow.net/feeds/question/51643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51643/monomorphisms-in-functor-categories Monomorphisms in functor categories Ralph 2011-01-10T12:46:27Z 2011-01-10T14:04:09Z <p>Let $I$ be a directed category and let $A$ be the category of $R$-modules ($R$ any ring). I'm trying to understand why the direct limit functor $$\varinjlim_{I}: A^I \to A$$ is exact. Since it has a right adjoint it's sufficient to show that it preserves monomorphisms. Let $\alpha: F \to G$ be a monomorphism in $A^I$. The proofs in the literature I know of (Weibel: "Introduction to homological algebra", 2.6.15 or Eisenbud: "Commutative Algebra", A6.4) seem to use the following as definition for $\alpha$ being mono: </p> <blockquote><p> $\alpha(i): F(i) \to G(i)$ is a monomorphism in $A$ for each $i \in obj(I).$ $\hspace{27pt}$ $\hspace{27pt}$ $(*)$ </p></blockquote> <p>It's easy to see that $(*)$ implies that $\alpha$ is mono in the usual sense (e.g. if $\beta: H \to F$ is a homomorphism in $A^I$ such that $\alpha \beta = 0$ then $\beta = 0$). </p> <p>Does anyone know if $(*)$ is equivalent to this definition of a monomorphism ? </p> <p>I was only able to settle the following special case: </p> <blockquote><p> Let $A$ be an abelian category and $I$ a small category such that for all $i, j \in obj(I)$: <ul> <li>$Hom_I(i,i) = \lbrace id_i \rbrace$ </li> <li>$Hom_I(i,j) \neq \emptyset \implies Hom_I(j,i) = \emptyset \hspace{5pt} (i \neq j)$</li> </ul> Then $\alpha$ above is mono iff $(*)$ holds. </p></blockquote> http://mathoverflow.net/questions/51643/monomorphisms-in-functor-categories/51647#51647 Answer by Martin Brandenburg for Monomorphisms in functor categories Martin Brandenburg 2011-01-10T13:52:20Z 2011-01-10T13:52:20Z <p>$A^I$ is an abelian category. Namely, the direct sums, kernels and cokernels may be constructed pointwise. In particular, $\alpha$ is mono iff $\ker(\alpha)=0$. But a functor $I \to A$ vanishes iff it vanishes pointwise. Thus $\alpha$ mono iff all $\alpha(i)$ mono.</p> http://mathoverflow.net/questions/51643/monomorphisms-in-functor-categories/51648#51648 Answer by Finn Lawler for Monomorphisms in functor categories Finn Lawler 2011-01-10T13:53:46Z 2011-01-10T13:53:46Z <p>The answer is yes if $A=R\mbox{-Mod}$ has pullbacks (which I'm pretty sure it does). See Tom Leinster's <a href="http://mathoverflow.net/questions/17953/can-epi-mono-for-natural-transformations-be-checked-pointwise/17977#17977" rel="nofollow">answer</a> to a similar question.</p> http://mathoverflow.net/questions/51643/monomorphisms-in-functor-categories/51650#51650 Answer by Buschi Sergio for Monomorphisms in functor categories Buschi Sergio 2011-01-10T14:04:09Z 2011-01-10T14:04:09Z <p>An Abelian category where filtrant colimts are exats is called a (abelian) Grothendieck category, no all Abelan categories are Grothendieck. Of course R-Mod is Grothendieck, because the forgetful functors U: R-Mod -> Set create limits (then preserve and reflexct Monomorphisms) and U create filtrant colimits, just because these (in Set) are coherent with finite products and algebraic structures. </p>