Why do filtered colimits commute with finite limits?

It's not hard to show that this is true in the category Set, and proofs have been written down in many places. But all the ones I know are a bit fiddly.

Question 1: is there a soft proof of this fact?

For example, here's a soft proof of the fact that filtered colimits in Set commute with binary products. If $J$ is a filtered category, and $R,S:J\to$ Set are functors, then

$$colim_{j\in J} R(j)\times colim_{k\in J} S(k) \cong colim_{j\in J} colim_{k\in J} R(j)\times S(k)$$ $$\cong colim_{(j,k)\in J\times J} R(j)\times S(k) \cong colim_{j\in J} R(j)\times S(j)$$

where the first isomorphism uses the fact that Set is cartesian closed, so that the functors $X\times-$ and $-\times X$ are cocontinuous; the second isomorphism is the "Fubini theorem"; and the third isomorphism follows from the fact that the diagonal functor $\Delta:J\to J\times J$ is final.

Is there some way to extend this to deal with equalizers and/or pullbacks? (The case of the terminal object is easy.)

For the sort of person who'd rather just prove the fact directly (which after all is not that hard), it's worth pointing out that this proof works not just in Set but for any cartesian closed category with filtered colimits. It works without knowing how to construct colimits in Set.

So another way to ask my question might be

Question 2: what is a class of categories in which you can prove that filtered colimits commute with finite limits (without first proving that this is true in Set)?

So yes, I know that the commutativity holds in any locally finitely presentable category, but the only proofs of this I know depend on the fact that it is true in Set.

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Thought I'd mention that a proof that filtered colimits commute with, say, equalizers, must use something more than what you've used for finite limits. Indeed, all you really used was that $J$ was "sifted" (the diagonal map is cofinal). But sifted colimits do not commute with equalizers in general. – Dylan Wilson Mar 10 '13 at 1:23
Thanks, Dylan. I agree that J being filtered will have to be used. But what properties of Set should be used? – Steve Lack Mar 11 '13 at 23:55

In the Elephant, Theorem B2.6.8 shows that finite limits commute with filtered colimits in $\mathsf{Set}$ using arguments that can apparently be internalized to any $\mathcal{S}$ which is Barr-exact with reflexive coequalizers. Let's call such a category good.

I expected Johnstone's proof to be a straightforward internalization of the proof found, say, in Mac Lane. But in fact he relies on reducing preservation of pullbacks to preservation of binary products, as Buschi Sergio attempted to do in his answer. Johnstone reduces from statement 1 to statement 2 as follows:

1. For any good category $\mathcal{S}$, and any $\mathbb{C} \in \mathrm{Cat}(\mathcal{S})$ which is internally filtered, the functor $\varinjlim: [\mathbb{C},\mathcal{S}] \to \mathcal{S}$ preserves pullbacks.

2. For any good category $\mathcal{S}$, and any $\mathbb{C} \in \mathrm{Cat}(\mathcal{S})$ which is internally filtered, the functor $\varinjlim: [\mathbb{C},\mathcal{S}] \to \mathcal{S}$ preserves binary products.

Johnstone proves statement (2) directly, but if we're willing to assume that $\mathcal{S}$ is cartesian closed, then I suppose statement (2) will follow in a more conceptual manner by internalizing the argument from the question statement.

Johnstone proves statement (1) from statement (2) as follows; I'll omit the word internal" a lot. Think of $[\mathbb{C},\mathcal{S}]$ as the category of discrete opfibrations over $\mathbb{C}$. Consider a pullback $\mathbb{G} \times_{\mathbb{F}} \mathbb{H}$ over the discrete opfibration $\mathbb{F} \to \mathbb{C}$. Then $\mathbb{G}$ and $\mathbb{H}$ can be regarded as discrete opfibrations over $\mathbb{F}$ in the slice category $\mathcal{S}/\pi_0 \mathbb{F}$, and $\mathbb{G}\times_\mathbb{F} \mathbb{H}$ is their product as such. Now, $\mathbb{F}$ is weakly filtered (meaning its connected components are filtered) over $\mathbb{S}$ by Johnstone's Lemma B2.6.7 (being a discrete opfibration over a filtered category), so it is filtered internally to $\mathbb{S}/\pi_0\mathbb{F}$ by Johnstone's Corollary B2.6.6. Hence, since $\mathcal{S}/\pi_0\mathbb{F}$ is again a good category, we can apply statement (2) to deduce that the product $\mathbb{G}\times_\mathbb{F} \mathbb{H}$ is preserved by the colimit functor $\varinjlim:[\mathbb{F},\mathcal{S}/\pi_0\mathbb{F}] \to \mathcal{S}/\pi_0\mathbb{F}$: $\varinjlim(\mathbb{G}\times_\mathbb{F} \mathbb{H}) \cong \varinjlim(\mathbb{G}) \times \varinjlim(\mathbb{H})$. When we apply the forgetful functor $\mathcal{S}/\pi_0\mathbb{F} \to \mathcal{S}$ to this isomorphism, colimits are preserved and products become pullbacks over $\pi_0 \mathbb{F}$, so it says

$\varinjlim(\mathbb{G}\times_\mathbb{F} \mathbb{H}) \cong \varinjlim(\mathbb{G}) \times_{\pi_0 \mathbb{F}} \varinjlim(\mathbb{H}) = \varinjlim(\mathbb{G}) \times_{\varinjlim( \mathbb{F})} \varinjlim(\mathbb{H})$

as desired. Note that in order to use the soft proof of (2), though, we need the slice category of $\mathcal{S}$ to be cartesian closed, i.e. we need $\mathcal{S}$ to be locally cartesian closed in addition to being good.

Some thoughts:

• In the direction of making this more self-contained, it looks like this proof could be stripped down to avoid reliance on internal logic if we just want it to apply when $\mathcal{S} = \mathsf{Set}$ -- although it looks like we will still have to think about categories internal to slices of $\mathsf{Set}$, this shouldn't be too bad. I'm not sure how soft" this is, though.

• In the direction of looking for maximum generality, this theorem identifies a nice class of categories where an internal version of finite limits and filtered colimits commute. But Question 2 asked for a nice class of categories where honest-to-goodness external finite limits commute with filtered colimits. I'm less sure about how to use this theorem to identify such a class. If $\mathcal{S}$ admits a geometric morphism to $\mathsf{Set}$ (or something along these lines), then ordinary small categories can be turned freely into internal categories in $\mathcal{S}$. Would such a functor also turn discrete opfibrations into discrete opfibrations? And would it preserve notions of limit and colimit? These are change-of-base questions that someone out there surely knows...

• It sure would be nice to modify this proof or find another proof which explicitly exploits the definition of filteredness of $\mathbb{C}$ which says that the diagonal functor $\Delta: \mathbb{C} \to [\mathbb{I},\mathbb{C}]$ is final for every finite $\mathbb{I}$.

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I'm still not sure whether this theorem implies that finite limits commute with filtered colimits externally in an arbitrary (Grothendieck) topos. I'm not sure whether that's even true! – Tim Campion Oct 28 '14 at 11:00
It is true. The usual proof uses the fact that sheafification preserves finite limits as well as filtered colimits. – Zhen Lin Oct 28 '14 at 16:51

For a generalization to pullback we have to proof that $colim_i X_i\times_{Y_i} B_i \cong X\times_YB$ (where $X, Y, B$ are the respective colimits). Because $I$ is filtred the triple diagonal $I\to I\times I\times I$ is final and we can make this colimit partially, then we can do the colimit in the $Y_i$ before.

Then we have to prove that $colim_i X_i\times_Y B_i \cong X\times_YB$ .

Then is enought show that the pullback of any colimit is still a colimit, and then with the some "soft proof" argumentations you done.

Is enought to show that:

give a $f: X\to Y$ and a cocone $B_i \to Y$ with $I$ a small category (no necessarly filtred), with a colimit $B_i\to B$ and the natural arrow $B\to Y$. Then the pullback with $f$: $B_i\times_Y X \to B\times_Y X$ is a colimit.

this is true if the pullbach funtor $(X, f)^\ast: \mathcal{C}\downarrow Y\to \mathcal{C}\downarrow X$ is a left adjoint, and then is cocomplete.

This is as said that $\mathcal{C}$ is locally-cartesian-closed.

This is true in any topos, and this property is a specific and profound aspect of topoi and their internal logic.

We can observe that in my above argomentation $I$ need not be filtred, but for $I$ no filtred the diagonal $I\to I\times I$ could be no final.

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Thanks, Buschi. I certainly agree that if the category is locally cartesian closed then we have $colim_i(X_i\times_Y B_i)\cong (colim_i X_i)\times_Y (colim_i B_i)$: this is the same argument I gave applied to the slice category ${\mathcal C}\downarrow Y$. But I don't follow the first paragraph. For one thing, the triple diagonal $I\to I\times I\times I$ does not seem to be directly relevant. You can form the product $X_i\times B_j$ for any $i$ and $j$, but the pullback $X_i\times_{Y_k} B_j$ only makes sense if we have first chosen maps $i\to k$ and $j\to k$. – Steve Lack Mar 2 '11 at 22:27
Steve LAck, yes you are right, in the first part I wrong (I seem it simple, I was shallow). I'm thinking about. – Buschi Sergio Mar 3 '11 at 9:44