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Stream: learning: questions

Topic: Associativity of limits = limits commute with limits?

Ralph Sarkis (Nov 01 2022 at 16:03):

Prologue: The procrastination gods set upon me, so sorry if this is another simple question where I am not seeing the solution. TL;DR: Is there a general notion of associativity of limits? If so, is it related to the fact that limits commute with limits?

Binary products are commutative in the sense that $A\times B \cong B \times A$. We can prove this very easily, but it is also readily apparent in the definition that $A\times B$ and $B\times A$ satisfy the same universal property (even more so when defining arbitrary products). Similarly, equalizers are commutative in the sense that $\mathrm{eq}(f,g) \cong \mathrm{eq}(g,f)$, and pullback are commutative $f\times_B g \cong g \times_B f$ (with the notation mentioned here).

We can generalize this as follows. For any isomorphism $\mathrm{swap}: \mathbf{J} \to \mathbf{J}$, we have the following adjunctions.
image.png

We can show that the top functor is isomorphic to $\lim_{\mathbf{J}}$ hence it is right adjoint to the diagonal functor. Intuitively, this means that you can scramble (with an isomorphism) your diagram before taking the limit and it will not change the outcome. Letting $\mathrm{swap}$ be the functor that swaps the two objects in the two object diagram, we get back the commutativity of products.

Question: Can we do the same thing for associativity?
Associativity for products means $(A\times B) \times C \cong A \times (B\times C)$. This can be shown using the fact that limits commute with limits and that the terminal object $\mathbf{1}$ is neutral wrt $\times$:

$$(A\times B) \times C \cong (A\times B) \times (1 \times C) \cong (A\times 1) \times (B\times C) \cong A\times (B\times C)$$

Taking equalizers is also associative, but I have not seen this before, so maybe my way of doing it is not as straightforward as it should be. Given three morphisms $f,g,h : A\to B$, the diagram below shows two ways of obtaining the equalizer of the three morphisms using binary equalizers only.
image.png

This looks really close to something that would follow from limits commuting with limits (especially if you write $f= \mathrm{eq}(f)$ and $h= \mathrm{eq}(h)$, but I am not seeing how it actually goes.
Question: Is the fact that limits commute with limits a generalization of associativity of products and equalizers?

For pullbacks, Borceux calls the pasting lemma "associativity of pullbacks", and I trust this post to reconcile this with a more intuitive notion of associativity.
Question: Is the fact that limits commute with limits a generalization of the pasting lemma? Of associativity of limits in general?

Mike Shulman (Nov 01 2022 at 16:08):

I think the right thing to generalize is "unbiased associativity":
$(A_{11} \times \cdots \times A_{1k_1}) \times \cdots \times (A_{n1}\times\cdots\times A_{nk_n}) \cong A_{11} \times\cdots \times A_{nk_n}.$
This can be seen as a specialization of the general fact that for any functor $p:\mathbf{K} \to \mathbf{J}$, we have $\lim_{\mathbf{J}} \circ \mathrm{Ran}_p \cong \lim_{\mathbf{K}}$. In particular, if $p$ is a fibration, then $\mathrm{Ran}_p$ can be computed by taking limits over the fiber category; and in even more particular, if $\mathbf{J}$ and $\mathbf{K}$ are finite and discrete, one recovers the above unbiased associativity for products.

Ralph Sarkis (Nov 01 2022 at 19:04):

That's interesting, but I am not comfortable enough with Kan extensions to understand how this applies to other limits. Your general fact seems to even generalize my commutativity fact as you are saying that the top functor in the following diagram is isomorphic to $\lim_{\mathbf{K}}$
image.png

I'll try for equalizers. Let us have the following two categories, and define $p$ by $p(k_1)= p(k_2) = j_1$ and $p(k_3) = j_2$.
image.png

What is the action of $\mathrm{Ran}_p$? If I have a functor $F: \mathbf{K} \to \mathbf{C}$ that I identify with three parallel morphisms $f,g,h : A\to B$. I guess the functor $\mathbf{Ran}_pF: \mathbf{J} \to \mathbf{C}$ should be identified with the two morphisms in the equalizer I took for $D$ in my second diagram above, and the components of the natural transformation $\mathrm{Ran}_pF \circ p \Rightarrow F$ are all written in this "commutative" diagram (only parts of it commute, precisely the parts that make the transformation natural).
image.png

Now, any other functor (that we identify with) $m_1,m_2: A' \to B'$ with a natural transformation whose components are $\phi_1:A' \to A$ and $\phi_2: B' \to B$ factorizes as follows, where $!$ is the unique morphism coming from the fact that $\phi_1$ equalizes $f$ and $g$ (this follows from the naturality).
image.png

Cool! I think I am satisfied with that. Let's say I won't do pullbacks because they are just products in a slice category.

Ralph Sarkis (Nov 01 2022 at 19:05):

Mike Shulman said:

This can be seen as a specialization of the general fact that for any functor $p:\mathbf{K} \to \mathbf{J}$, we have $\lim_{\mathbf{J}} \circ \mathrm{Ran}_p \cong \lim_{\mathbf{K}}$.

So is this fact related to limits commuting with limits?

Mike Shulman (Nov 01 2022 at 19:36):

Yes, this fact is like the version of the fubini theorem that says $\int_J \int _K = \int_{J\times K}$, and then you compose it with the other version of itself to get $\int_J \int_K = \int_K \int_J$ which is limits commuting with limits.