# Abelian categories, axioms AB5 and AB5* and incompatability

This is a homework exercise, so please don't post full solutions to the question below.

Grothendieck (I believe) introduced several axioms an abelian category A voluntarily could satisfy. In particular, we have the two following:

1. AB5 A is cocomplete and filtered colimits of exact sequences are exact.

2 AB5* A is complete and filtered inverse limits of exact sequences are exact.

Then my problem is the following:

A.4.7 (Weibel) Show that if $A\neq 0$, then A cannot satisfy both axiom AB5 and AB5*. Hint: consider $\oplus A_i \to \prod A_i$.

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I have stared at this problem for a long time but no ideas have (nor much intuition for filtered limits) been born. If anyone has any helpful explanations or wonderful hints, I'd be really grateful. Thanks.

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The axioms are not voluntary... There are categories which try hard and hard but do not manage! – Mariano Suárez-Alvarez Mar 8 '11 at 2:05
It is really hard to say anything more than Weibel without giving the solution away... What kind of morphism is $\bigoplus A_{i} \to \prod A_{i}$? – t.b. Mar 8 '11 at 5:59
It is a monic (thinking in elements, it is the obvious injection, I presume). So I guess I am to take some limit of this, but I don't see what limit. Precisely: I don't think I see how "filtered" fits in the picture. – Fredrik Meyer Mar 8 '11 at 6:08
You can say more (use self-duality of the hypotheses!). But why exactly is it a monic? Take the finite sums $A_{i_1} \oplus \cdots \oplus A_{i_{n}}$ where $\{i_{1},\ldots,i_{n}\}$ runs through the finite subsets of $I$. These form a filtered set and each of those sums injects into $\prod A_{i}$, hence so does their colimit $\bigoplus A_{i}$ by AB5. Now dualize. – t.b. Mar 8 '11 at 8:34