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$?


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 enter image description here

The above statement now is that the transfinite compositions of the left and right column coincide up to a canonical isomorphism.

  • 1
    $\begingroup$ I think there's more than just a notational problem here. How do you define infinitely iterated colimits? $\endgroup$
    – Zhen Lin
    Nov 3, 2014 at 8:08
  • $\begingroup$ Talking of iteration of limits, you may be interested in (a categorical version of) mathoverflow.net/questions/12211/… and mathoverflow.net/questions/29427/… $\endgroup$ Nov 3, 2014 at 8:30
  • $\begingroup$ @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" $\endgroup$
    – fosco
    Nov 3, 2014 at 10:59
  • 1
    $\begingroup$ 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 $\varinjlim_n X_n$. $\endgroup$
    – Zhen Lin
    Nov 3, 2014 at 11:48
  • 2
    $\begingroup$ 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). $\endgroup$ May 30, 2016 at 18:42


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