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 $\oplus A_i \to \prod A_i$.


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.

  • $\begingroup$ The axioms are not voluntary... There are categories which try hard and hard but do not manage! $\endgroup$ – Mariano Suárez-Álvarez Mar 8 '11 at 2:05
  • $\begingroup$ It is really hard to say anything more than Weibel without giving the solution away... What kind of morphism is $\bigoplus A_{i} \to \prod A_{i}$? $\endgroup$ – t.b. Mar 8 '11 at 5:59
  • $\begingroup$ 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. $\endgroup$ – Fredrik Meyer Mar 8 '11 at 6:08
  • 2
    $\begingroup$ 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. $\endgroup$ – t.b. Mar 8 '11 at 8:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.