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 $i, j \in obj(I)$:

Then $\alpha$ above is mono iff $(*)$ holds.

- $Hom_I(i,i) = \lbrace id_i \rbrace $
- $Hom_I(i,j) \neq \emptyset \implies Hom_I(j,i) = \emptyset \hspace{5pt} (i \neq j)$

dualis wrong. Filtered limits of short exact sequences need not be exact, see en.wikipedia.org/wiki/Mittag-Leffler_condition – Theo Buehler Jan 10 '11 at 14:40