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:
$\alpha(i): F(i) \to G(i)$ is a monomorphism in $A$ for each $i \in obj(I).$ $\hspace{27pt}$ $\hspace{27pt}$ $(*)$
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 $\beta = 0$).
Does anyone know if $(*)$ is equivalent to this definition of a monomorphism ?
I was only able to settle the following special case:
Let $A$ be an abelian category and $I$ a small category such that for all
$i, j \in obj(I)$:
- $Hom_I(i,i) = \lbrace id_i \rbrace $
- $Hom_I(i,j) \neq \emptyset \implies Hom_I(j,i) = \emptyset \hspace{5pt} (i \neq j)$
Then $\alpha$ above is mono iff $(*)$ holds.
Best Answer
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.