Does the limit(colimit) have transitivity?

Question

For example, in $Set$:

The limit of $A$ and $B$ is $A \times B$;

The limit of $A \times B$ and $C$ is ($A \times B) \times C$;

The limit of $A$, $B$ and $C$ are $A \times B \times C$.

We know that ($A \times B) \times C \cong A \times B \times C$.

My question is:

Can this conclusion be generalized to any limit/colimit than product/coproduct?

I searched a conclusion on nlab, it give me a conclusion.

https://ncatlab.org/nlab/show/preserved+limit#Examples

limits preserve limits

https://ncatlab.org/nlab/show/limits+commute+with+limits

limits commute with limits

It seems to tell me Yes, but its explanation is a bit complicated for me. Since I am not a mathematician, so I want to confirm this fact.

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Edit for making this question clearer.

Given a category $C$ and two diagrams $D : I \to C$ and $E : J \to C$, suppose the image of $I$ and $J$ are not overlapped and we know that $D$ and $E$ have limits (i.e. $\lim_D$ an $lim_E$). Now I can take the product of the two apex of $\lim_D$ an $\lim_E$ (suppose the product exists).

The question is:

Is there any relationship between $\lim_D \times \lim_E$ and $\lim_{D+E}$ ?

For example,

1. Does the $\lim_{D+E}$ exist, if $\lim_D \times \lim_E$ exist?

2. Does the apex of $\lim_D \times \lim_E$ isomorphic to $\lim_{D+E}$? I'm sorry, since I don't know how to compare two limits equipped with different cones, so I can only compare apex here. (In this case, the diagram of $\lim_{D+E}$ is $D+E$ on $C$ and the diagram of $\lim_D \times \lim_E$ is two objects.)

Thanks.

Answer 1

The answer to the specific question is yes.

More precisely :

Let $I,J$ be two categories, $F:I\to C, G: J\to C$ two functors. Let $I+J$ be the coproduct of $I,J$, that is, a category which consists of one copy of $I$ and one disjoint copy of $J$ with no morphisms inbetween; then let $K: I+J\to C$ be obtained from $F,G$ by the universal property of the coproduct. Assume $\lim F$ and $\lim G$ exist. Then $\lim K$ exists if and only if $\lim F\times \lim G$ exists, in which case there is a canonical isomorphism between the two.

I won't dwell on what "canonical" means here, but hopefully it's clear from the statement (and/or the proof).

Proof : For notations, for $i\in I, p_i: \lim F\to F(i)$ is the canonical morphism, and similarly $q_j : \lim G\to G(j)$ for $j\in J$

Assume $P= \lim F\times \lim G$ exist, with projections $p$ to $\lim F$ and $q$ to $\lim G$. Then define for $k\in I+J$, $r_k : P\to K(k)$ by $p_i\circ p$ if $k=i\in I$, $q_j\circ q$ if $k=j\in J$.

I claim that this is a cone $P\to K$. To see this, note that any morphism in $I+J$ is either in $I$, or in $J$, in which case the cone property follows from that of $\lim F$, or $\lim G$.

Therefore $P$ is a cone. Moreover it is universal : let $C\to K$ be a cone. Then the restriction to $I$ is a cone $C\to F$, which induces a morphism $a : C\to \lim F$, and similarly by restricting to $J$ we get a map $b :C\to \lim G$. Using the universal property of products, we get $C\to P$. It's easy to check that it's a factorization of my cone.

It's also easy to check that any such factorization must be of this form, and therefore by uniqueness of the maps $C\to \lim F$ and $C\to \lim G$, this factorization is unique.

I'll let you try to prove the converse, it's essentially the same kind of reasoning, not too hard - if you have trouble with it, don't hesitate to ask.

Warning: here we saw that you if you have a category $L$ which can be written as $I+J$, then you can express the limit of $L$ as function of the limits on $I$ and $J$.

In general, it is difficult to relate the limit along a given category and the limits along categories that appear in a "decomposition" of said category - you have to know, in a good way, how those categories interact. Here, $I$ and $J$ essentially don't interact, which is what makes the description really easy.