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I was looking at this question about a "soft proof" of the fact that finite limits (shape $I$) commute with filtered colimits (shape $J$) in Set, using only the fact that the diagonal $J \to J^I$ is final.

If we consider the simple case of intersection and unions of sets $A_{i,j}$, we have the formula

$$\bigcap_{i \in I}\bigcup_{j \in J} A_{i,j} = \bigcup_{f : I \to J} \bigcap_{i \in I} A_{i,f(i)}$$

where $f : I \to J$ is a "skolemization" of the existential quantifier on the left hand side (This requires AC for infinite $I$). I am wondering if in the same vein, a limit-of-colimits can be replaced by a colimit-of-limits over a diagram category.

That is, do we have for every diagram $A: I \times J \to \mathbf{Set}$ ($I$ possibly finite) that $$\mathrm{lim}_{i \in I}\mathrm{colim}_{j \in J}\; A(i,j) \cong \mathrm{colim}_{F \in J^I} \mathrm{lim}_{i \in I}\; A(i,F(i))$$

In this case, using the finality of $I \to J^I$ we'd obtain the desired commutation

$$\mathrm{lim}_{i \in I}\mathrm{colim}_{j \in J}\; A(i,j) \cong \mathrm{colim}_{j \in J} \mathrm{lim}_{i \in I}\; A(i,j)$$

So the formula should hold for $I$ finite $J$ filtered, and I've checked it for $J$ discrete as well if I'm not mistaken. In general, going from the lhs to the rhs requires to build up a skolemizing diagram $F$ not only on objects but on morphisms as well, which seems to involve some complicated choice. Is a formula like the above known, or is my intuition off somewhere?

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    $\begingroup$ I'm wondering if something like this might also be relevant to distributivity of limits and colimits. $\endgroup$ – Tim Campion Feb 5 at 14:49
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    $\begingroup$ More precisely, the $g,h$ factorization which appears in Section 2 on the above nlab page is the same factorization of the commutativity comparison map that you're using. I've also asked this related question. Reid and Brian's thoughts make me want to think that in the $\infty$-category of spaces, all limits distribute over all colimits. I think it would follow from the discussion in Dario's question that every $\infty$-categorical doctrine is sound, which sounds like I'm wrong, but I'm intrigued. $\endgroup$ – Tim Campion Feb 5 at 19:47
  • $\begingroup$ I naively assumed commutativity and distributivity of limits&colimits were equivalent terminology, but distributivity is indeed well-named and gives me a good notion to look into further. Thanks $\endgroup$ – Dario Stein Feb 10 at 10:24
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This isn't true in general. Take $I = BG$ and $J = BH$ to be one-object groupoids, so that $A(i, j)$ becomes a set $A$ with commuting actions of $G$ and $H$. The left hand side is obtained by taking the $H$-orbits of $A$, and then the $G$-fixed points of the result. The right hand side isn't as easy to describe precisely, but it's some quotient of the coproduct over all group homomorphisms $f : G \to H$ of the elements of $A$ fixed by the action of $(g, f(g))$ for every $g \in G$.

Take $G = \mathbb{Z}/2$, $H = \mathbb{Z}$ and let $A = \mathbb{Z}/2$ with action of both $G$ and $H$ given by addition (mod 2). Then $A$ has a single $H$-orbit, which is (obviously) fixed by $G$, so the left hand side has a single element. On the other hand, the only group homomorphism $f : \mathbb{Z}/2 \to \mathbb{Z}$ is the zero map, and there aren't any elements of $A$ fixed by the action of $(g, 0)$ for $g = 1 \in \mathbb{Z}/2$, so the right hand side is empty.

What went wrong in this example? Well, let's try to construct an isomorphism between the two sides anyways. Given an element $x$ on the left, we should try to construct a "corresponding" group homomorphism $f : G \to H$ (not necessarily uniquely determined). The element $x$ is an $H$-orbit of $A$ which is fixed by $G$. Let's pick an element $a \in A$ of this orbit. Then for each $g \in G$, since $ga$ generates the same orbit as $a$, there must be some $h \in H$ with $(g, h)a = a$. We would like to send $x$ to the rule $f$ which maps $g$ to such an $h$. However, $h$ is only well-defined as an element of some quotient of $H$, namely the quotient by the subgroup which fixes $a$ (or $ga$, since the actions commute). So we obtain a group homomorphism from $G$ to a quotient of $H$, but it might not lift to $H$ itself.

One thing I'm not sure about is what happens in the $\infty$-category of spaces, which generally has better exactness properties since there is no set-truncation in the construction of colimits. If the isomorphism holds there, then one could recover the version for sets when $J$ is filtered, since filtered colimits of sets are also homotopy colimits.

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    $\begingroup$ It seems the analogous statement in the $\infty$-category of $\infty$-groupoids is true. One can see this by viewing $A$ as a left fibration $\int A \to I \times J$. Then both sides of the putative equivalence describe sections of the composite $\int A \to I\times J \to I$. (The left hand side is directly just sections, whereas the right hand side describes first constructing a section of $I \times J \to I$, and then lifting that.) unfortunately, I'm not sure where the precise details for this can be found. $\endgroup$ – Brian Shin Feb 5 at 15:23
  • $\begingroup$ @Reid Barton: Thanks for the nice counterexample. I'm very curious to learn more about the $\infty$-categorical perspective, and the connection to filtered limits. $\endgroup$ – Dario Stein Feb 5 at 19:16
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    $\begingroup$ Ah, I think my earlier comment may need some modifications if $I$ and $J$ are not themselves $\infty$-groupoids. $\endgroup$ – Brian Shin Feb 6 at 4:54

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