Let $A_n, B_n, C_n$ be directed systems in some abelian category. Denote by $A \times_C B$ the fibre product of $A$ and $B$ over $C$.

Is it true that $(\varprojlim A_n) \times_{\varprojlim C_n} (\varprojlim B_n) = \varprojlim (A_n \times_{C_n} B_n)$, if $\varprojlim (A_n \rightarrow C_n)= (A \rightarrow C)$ and similarly for $B$?

I know that "morally" limits preserve limits, but I don't know the exact statement I need here (i.e. limits in which categories?)

Edit: Analogously, what can we say after replacing filtered limits with filtered colimits? Edit2: I think in the case of filtered colimits, I just checked the universal property of a colimit. In case I didn't make a mistake, the statement holds. The other case (filtered limits), and arbitrary colimits/limits are not as important for my purposes, but if you have a nice statement, please feel free to post it.

  • 1
    $\begingroup$ Limits commute with limits in the sense you suggest (always). The story about limits and colimits is more complicated. $\endgroup$ – Zhen Lin May 24 '14 at 22:17
  • $\begingroup$ Filtered colimits commute with finite limits in $\text{Set}$ and in categories that are sufficiently similar to $\text{Set}$. $\endgroup$ – Qiaochu Yuan May 24 '14 at 22:58
  • $\begingroup$ @Qiaochu Yuan Are categories of modules over rings sufficiently similar to $Set$? Also, I'd be happy to see a reference? Is there a statement in MacLane? Thank you! $\endgroup$ – user110071 May 24 '14 at 23:31
  • $\begingroup$ @user: yes. Any category of models of a Lawvere theory works. I don't know a reference; the statement for $\text{Set}$ at least should be somewhere in Borceux. See qchu.wordpress.com/2013/06/23/… for some details (I don't quite prove this statement but it can be inferred from things I do prove if you believe the statement for $\text{Set}$). $\endgroup$ – Qiaochu Yuan May 24 '14 at 23:38

Limits always commute with limits. Here is a precise statement. Let $J_1, J_2$ be two diagram categories and let $F : J_1 \times J_2 \to C$ be a diagram in a category $C$. Then whenever the limits

$$\lim_{j_1 \in J_1} \lim_{j_2 \in J_2} F(j_1, j_2)$$


$$\lim_{j_2 \in J_2} \lim_{j_1 \in J_1} F(j_1, j_2)$$

exist, they are canonically isomorphic because they are both canonically isomorphic to the limit $\lim_{(j_1, j_2) \in J_1 \times J_2} F(j_1, j_2)$ of $F$ itself. For a discussion see Section 2.12 in Borceux. (To be totally precise, those inner limits are taking place in the category of functors $J_1 \to C$ resp. the category of functors $J_2 \to C$, but limits in functor categories are always computed pointwise so I think the abuse of notation is forgivable. You can think of this as a statement about adjoints composing.)

The description of when limits commute with colimits is more complicated. Perhaps the most important case is that filtered colimits always commute with finite limits in $\text{Set}$, and hence in any category $C$ equipped with a faithful functor $U : C \to \text{Set}$ preserving and reflecting filtered colimits and finite limits; in particular, categories of modules have this property. For a discussion see Section 2.13 in Borceux.

The corresponding statement for arbitrary abelian categories is false; in particular, there's no reason for it to be true in $\text{Ab}^{op}$. To write down a counterexample in $\text{Ab}^{op}$ means to find an example where cofiltered limits fail to commute with pushouts in $\text{Ab}$; as a special case this means to find an example where cofiltered limits fail to commute with cokernels.

Here is an explicit counterexample. Consider the cofiltered diagram of morphisms $p^n \mathbb{Z} \to \mathbb{Z}$ where $p$ is some prime. The cofiltered limit of the cokernels is the $p$-adic integers $\mathbb{Z}_p$, but the cokernel of the cofiltered limit is $\mathbb{Z}$ (the cofiltered limit itself is $\mathbb{Z} \xrightarrow{0} \mathbb{Z}$).

This phenomenon implies in particular that the cofiltered limit fails to be exact in general, and in some situations this means one has to take its derived functor, often denoted $\lim^1$.

  • $\begingroup$ Borceux actually does not prove the first statement in the way I suggest, so here is the proof I have in mind: recall that taking limits of a diagram of shape $J$ is (pointwise) right adjoint to the diagonal functor $C \to C^J$. The diagonal functor $C \to C^{J_1 \times J_2}$ factors in two ways, first through $C^{J_1}$ and second through $C^{J_2}$. Now use the fact that adjoints compose. $\endgroup$ – Qiaochu Yuan May 25 '14 at 16:23

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.