I'm working through a proof of the existence of a canonical mapping

$$ \mu: \text{colim}_I \text{lim}_JH(i,j) \longrightarrow \text{lim}_J\text{colim}_IH(i,j) \tag{1} $$ induced by a cone $(\mu_i: \text{lim}_JH(i,-)\to \text{lim}_J\text{colim}_I H(i,j))_{i \in I}$ with $\mu_i$ defined as the map induced by the cone $(\mu_{ij})_{j \in J}$ defined as $$ \text{lim}_JH(i,-) \xrightarrow{\pi_j} H(i,j) \xrightarrow{\iota_i} \text{colim}_IH(-,j). $$

I have yet to show that these are in effect cones: what approach can be taken in order to prove this in the least 'chaotic' way possible? My attempts so far have involved an amount of arrows that I can hardly keep track of, so maybe there is an elegant solution lurking in the corners which I am failing to see.

  • $\begingroup$ In recommend not to start a war with cones, they are messy to keep track of, and since you already obviously assume the existence of limits and colimits over $I,J$, why not just use universal cones? you already got them for free. $\endgroup$ – Enkidu Jan 15 at 9:31
  • $\begingroup$ Would you mind elaborating? I am not seeing it. I do use universal cones when I talk about $(\iota_i\pi_j)_j$ but to relate this to universal cones the former should be a cone to begin with, which takes us back to the original question $\endgroup$ – Guido A. Jan 15 at 11:20
  • $\begingroup$ Well, all your objects are defined by universal properties over a diagram, I would suggest you forget the cones part and only look at the diagrams! since you have already canonical maps between them (the identities). respectively, you even have already the maps from the limits to the colimits. so you can just piece those together by universal properties. i.e. use the universal property of colimit and limit to build your map. $\endgroup$ – Enkidu Jan 15 at 11:26
  • $\begingroup$ But isn't that what I am using? A map out of the colimit is given by maps put of every object into $\text{lim}\text{colim}H$ such that they form a cone, and each of these is a map with a limit as a codomain, so it is universally characterized by maps into each object of it which form a cone. I have already defined this maps (in terms of the universal property of each inner (co) limit) as $\mu_{ij} = \iota_i \pi_j$. Now I want to prove that these are cones, in order to finish off the proof. Am I missing something? $\endgroup$ – Guido A. Jan 15 at 13:28
  • $\begingroup$ well, you are given that those are cones, by the universal property. $\endgroup$ – Enkidu Jan 15 at 13:33

I think the shortest way is to use two ingredients :

  1. the identification between functors $H:I\times J\to C$ and functors $I\to C^J$
  2. the functoriality of (co)limits.

Indeed if you think of $H$ as a functor $I\to C^J$, taking its colimit gives you a cone $$H(i,j) \xrightarrow{\iota_i} \text{colim}_IH(-,j).$$ in the functor category $C^J$. Then you can apply the functor $\lim_J$ on this cone to directly obtain the cone $$(\mu_i: \text{lim}_JH(i,-)\to \text{lim}_J\text{colim}_I H(i,j))_{i \in I}$$ in the category $C$. From there you get the canonical mapping $\mu$, as you say in your question.

  • $\begingroup$ Thanks for taking the time to answer. I have been looking at your post for a while but I can't seem to complete the missing details. What do you mean by applying the limit functor to both sides? Shouldn't we first have to construct some morphism in $C^J$ to which we could apply this functor to? $\endgroup$ – Guido A. Jan 15 at 17:15
  • $\begingroup$ @GuidoA. It is already in $C^J$, in fact. But I was probably too concise. I've edited a little bit, I hope it's better now. $\endgroup$ – Arnaud D. Jan 15 at 18:45
  • $\begingroup$ I was misinterpreting the former statement: I see how we can form maps $\mu_i$ in this way. But how can one say that this is a cone? $\endgroup$ – Guido A. Jan 15 at 20:28
  • $\begingroup$ @GuidoA. Cones are commutative diagrams (or natural transformation to a constant functor, if you prefer), so they are preserved by functors. $\endgroup$ – Arnaud D. Jan 15 at 21:26
  • $\begingroup$ This result confused me beyond the usual, I see it now! Thanks so much for your patience and the additional details :) $\endgroup$ – Guido A. Jan 15 at 22:04

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