# Do pullbacks commute with filtered limits in this sense?

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.

• Limits commute with limits in the sense you suggest (always). The story about limits and colimits is more complicated. – Zhen Lin May 24 '14 at 22:17
• Filtered colimits commute with finite limits in $\text{Set}$ and in categories that are sufficiently similar to $\text{Set}$. – Qiaochu Yuan May 24 '14 at 22:58
• @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! – user110071 May 24 '14 at 23:31
• @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 $\text{Set}$). – 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

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

and

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

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 $\lim^1$.

• 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. – Qiaochu Yuan May 25 '14 at 16:23