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Let $\mathbf C$ be an abelian category admitting projective limits. Let's consider the category whose objects are those of the form $$ 0\to A\to B\to C\to 0 $$ and whose morphisms are triples of morphisms of $\mathbf C$ such that the diagram $$ \begin{array}{ccccc} A&\hookrightarrow & B & \to & C & \newline \downarrow &&\downarrow && \downarrow \newline A' &\hookrightarrow & B' &\to & C' \end{array} $$ commutes in all its parts. Call this category $\boldsymbol\Sigma(\mathbf C)$.

How could one characterize the limits in $\boldsymbol\Sigma(\mathbf C)$? A little meditation shows that inverse systems are objects of the form $$ \mathcal E_i\colon 0\rightarrow A_i\rightarrow B_i\rightarrow C_i\to 0 $$ (every object in the sequence can be thought as an element in a separate inverse system), so the universal property of $\varprojlim_\mathbf J \mathcal E_i$, whatever it turns out to be, must be enjoyed by $$ \textstyle \varprojlim_\mathbf J \mathcal E_i : 0\to \varprojlim_\mathbf J A_i\to \varprojlim_\mathbf J B_i \to \varprojlim_\mathbf J C_i\to 0 $$ as soon as one looks to ${A_i},{B_i},{C_i}$ as three inverse systems.

What condition(s?) has(ve?) to be imposed on them to make sure the sequence is exact?

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    The title says "exact sequences", but in the body you do not require the short sequences to be exact. Could you please clarify this point?2010-10-26
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    Supposing for concreteness that you are looking at short exact sequences of modules over some ring, the condition that you need for the projective limit exact sequence to be exact is that the inverse system $\{A_i\}$ should satisfy the Mittag-Leffler condition. (This should be easy to read about by googling; it is also discussed in Hartshorne, surely in Weibel, and I imagine many other texts as well.)2010-10-26
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    Matt is right: http://en.wikipedia.org/wiki/Mittag-Leffler_condition#Derived_functors_of_the_inverse_limit2010-10-26

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Considering the category $\Sigma_R(C) $ of the exact seqences like: $0\to A\to B\to C$ this category is complete and the inclusion $\Sigma(C) \subset \Sigma _R(C)$ is coreflexive (the coreflector comes from the coker of the last arrow), then limits in $\Sigma(C) $ exist and are given by the limit in $\Sigma_R(C) $ followed by the coreflection.

When the inclusion $\Sigma(C) \subset \Sigma _R(C)$ preserves (directed) limits is the content of Mittag-Leffler's theorem.

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    Thanks, it is what I was looking for.2010-12-13