# Infinite Fubini rule for co/limits

It is a well known fact that given a functor $F\colon I\times J\to C$ then (when everybody exists) $$\varinjlim_I \varinjlim_J F\cong \varinjlim_J \varinjlim_I F\cong \varinjlim_{I\times J}F$$ Now, what if I have $F\colon \prod_\lambda I_\lambda\to C$? I have serious notational problems in stating the precise result, but "is something similar true"? Is it true that for any possible partition of $\Lambda$, and any possible order in which I saturate the various components of $F$, taking colimits of colimits of ... I obtain each time the same result $\varinjlim{}_{\prod C_\lambda} F$?

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Edit May 30, 2016: I finally have the time to expand this question showing you why I posted it years ago. I'm trying to prove or disprove the following statement:

Let $\Lambda$ be a set; let $T:\prod_{\lambda\in\Lambda}{\cal C}_\lambda^\text{op}\times {\cal C}_\lambda \to {\cal D}$ be a functor.

1. Let $w\colon \gamma\to \Lambda$ be a well-ordering; let $\int_{(c_\lambda),[w]}T$ denote the end of $T$ obtained following the ordering induced by the bijection between $\Lambda$ and $\gamma$.
2. Let $\sigma\colon \Lambda\to \Lambda$ be a bijection, and $w\colon \gamma \to \Lambda$ be a well-ordering; let $\int_{(c_\lambda), [\sigma]}T$ denote the end of $T$ obtained following the ordering $\gamma \xrightarrow{w} \Lambda \xrightarrow{\sigma}\Lambda$.
3. Let $(\Lambda_k)_{k\in K}$ be a partition of $\Lambda$ over a well-ordered set (this is not restrictive). Let $\int_{(c_\lambda),[K]}T$ be the end of $T$ obtained integrating $k$-wise over elements of $\Lambda_k$, via the canonical identification $\prod_{\lambda\in\Lambda}{\cal C}_\lambda^\text{op}\times {\cal C}_\lambda \cong \prod_{k\in K}\prod_{\lambda_k\in\Lambda_k}{\cal C}_{\lambda_k}^\text{op}\times {\cal C}_{\lambda_k}$

Each of these three objects exists if and only if the other two do, and they are all canonically isomorphic to the end $$\int_{(c_\lambda)\in \prod_\lambda{\cal C}_\lambda} T(c_\lambda, c_\lambda)$$ obtained via the identification $\prod_\lambda {\cal C}_\lambda^\text{op}\times {\cal C}_\lambda \cong \big(\prod_\lambda {\cal C}_\lambda\big)^\text{op} \times \prod_\lambda {\cal C}_\lambda$.

Here's an idea for the proof in the case $\Lambda = \mathbb{N}$ with its usual well-order, and $\sigma\colon \mathbb{N}\to \mathbb{N}$ is any bijection: by iteratively integrating over successive variables we obtain the diagram The above statement now is that the transfinite compositions of the left and right column coincide up to a canonical isomorphism.

• I think there's more than just a notational problem here. How do you define infinitely iterated colimits? – Zhen Lin Nov 3 '14 at 8:08
• Talking of iteration of limits, you may be interested in (a categorical version of) mathoverflow.net/questions/12211/… and mathoverflow.net/questions/29427/… – Pietro Majer Nov 3 '14 at 8:30
• @ZhenLin: considering that when you do $\lim_I\lim_J F$ you compose two functors, I think it's a transfinite composition. The claim would sound like "whatever I choose to well-order $\Lambda$, and then consider the transfinite composition of the chain I get, I obtain the same result" – Fosco Nov 3 '14 at 10:59
• It's really easy: if you have $f_n : X_n \to X_{n+1}$ then the transfinite composition $\cdots \circ f_2 \circ f_1 \circ f_0$ lands in $\varinjlim_n X_n$. – Zhen Lin Nov 3 '14 at 11:48
• I think that what you want is the Kan extension. It is an exercise in cofinality that the left Kan extension along the Grothendieck construction of a functor is the colimit computed fiberwise, and this seems to encode exactly what "Fubini" should mean (ditto for right Kan extensions and limits). – Denis Nardin May 30 '16 at 18:42