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Fix an index (small) category $I$. Let's say our category $\mathcal{C}$ has limits of type $I$. In this case, $\varprojlim_I:\mathcal{C}^I \to \mathcal{C}$ is a functor.

Let's say our category $\mathcal{C}$ also has limits of type $J$. In this case, $\varprojlim_J: \mathcal{C}^J\to \mathcal{C}$ is also functor.

How do these relate?

Let me clarify. Suppose now our category $\mathcal{C}$ is complete, i.e. has all (small) limits. Can we somehow define a functor $\varprojlim$ which takes any diagram of any type and outputs its limit?

Motivation: let $R$ be a ring (commutative for simplification). It's easy to formalize the statement that the isomorphism $A\otimes_R \bigoplus_{i\in I} B_i \simeq \bigoplus_{i\in I} (A\otimes_R B_i)$ of $R$-modules is "natural in $A$".

But can it be also natural in $\{B_i\}_{i\in I}$? Yes:

This isomorphism is a particular case of the more general statement $A\otimes_R \varinjlim_I F\simeq \varinjlim_I A\otimes_R F$, where $F:I\to R-\mathrm{Mod}$ is a functor. This isomorphism is natural also in the second variable, meaning that there is a natural isomorphism between two appropiate functors.

But what if we also let $I$ vary?

Going back to the particular case of the direct sum, there is this proposition (taken out of Rotman, Introduction to Homological Algebra, page 87):

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Now, this "naturality" involves also varying the index set $I$. How can this "naturality" really be expressed as a natural isomorphism between functors? Can it be generalized for arbitrary colimits?

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    To answer your very first question: yes, easily modulo set theoretic problems. One can define the category of "all small diagrams" in a very straightforward way. But $\varprojlim$ is not canonically a functor – even when you fix the diagram shape – in the sense that a sufficiently large axiom of choice is required to define it. To be totally canonical one has to instead look at anafunctors.2012-01-31
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    @Zhen, but in very concrete situations, it can be made into a functor without even knowing what «sufficiently large axiom of choice» even means: it is enough to be able to construct the limits in a concrete way, as in $\mathbf{Set}$, for example.2012-02-06
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    @Mariano: Indeed. But the hypothesis is usually of the form "the category $\mathcal{C}$ has limits of shape $J$", in which case a functorial choice of limits is not part of the hypothesis.2012-02-06
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    $\mathcal C^I=\text{Func}(I,\mathcal C)$, so for any $\phi\colon I\to J$ you get a functor $\Phi\colon \mathcal C^J \to \mathcal C^I\colon F\mapsto F\circ\phi$... maybe you're looking for some condition to have $$ \varinjlim_J F = \varinjlim_I( F\circ\phi) $$2012-01-31
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    I think I found what I was looking for, but I'm not really up to checking it: exercise 5 in page 111 of MacLane's CWM is called "limit as a functor on the comma category of all diagrams in $\mathcal{C}$", and I think it resolves the issue. If I ever bother in checking the details I'll post this as an answer. If someone is willing to do it, be my guest :)2012-06-07

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