Existence of limits of set-valued functors gives me weird limits

Question

I’m reading a theorem that says that a limit for an arbitrary set-valued functor always exists. It also say how to find it.

I will now illustrate a sample application of the theorem.

Given the following indexing category $I$

We would like to know what is the limit of the functor $X:I\to\mathbf{Set}$ here illustrated

Now, according to the theorem the limit would be isomorphic to a set of three elements, specifically $\operatorname{lim}_I X=\{(a_1,b_1),(a_2,b_1),(a_3,b_2)\}$.

To me it seems that the real limit is just any singleton with two functions to $A$ and $B$. This is because you can definitely find a mapping from the alleged “limit” to my limit, for example see the following

So i created some mock functions that don’t even touch $a_1$ and $a_2$. This way I can create a fine mapping from the supposed limit to my singleton in the slice category over $X$.

This theorem is very confusing and I don’t even understand why there is no distinction between the category of sets and the slice category over $X$. Is it possible that the limit is just a set? Shouldn’t it be a functor?

Answer 1

In any category, $\mathcal C$, the limit any diagram, $D:\mathbb I \to \mathcal C$ of shape $\mathbb I = \boxed{1\to 2}$ exists and its object part, i.e. the apex of the cone, is $D(1)$ up to isomorphism and the projection from that to $D(1)$ is $id$ which forces the projection from $D(1)$ to $D(2)$.

In other words, the limit in your example is $A$ up to isomorphism and indeed $\{(a_1,b_1),(a_2,b_1),(a_3,b_2)\}\cong A$.

Using a singleton doesn't work as a limit. While it (trivially) satisfies the uniqueness condition of the universal property describing limits, it does not satisfy the existence condition. The universal property of this limit is: $(L,p:L\to A)$ is a limit for $X$ iff for every $T$ and $f:T\to A$ there exists a unique $\bar f:T \to L$ such that $p\circ \bar f = f$. (There's no mention of $B$ or the map from $A\to B$, call it $h$, because once we have a map $f:T\to A$, we automatically have a map $h\circ f:T\to B$. Naively written, the universal property would also require a $q:L\to B$ but it would also require $q=p\circ h$. Similarly it would require a $g:T\to B$ in addition to $f$, but it would require $g=f\circ h$, so all the information is already held in the map to $A$.) To prove my earlier statement, clearly choosing $L=A$, $p=id_A$, and $\bar f =f$ satisfies the universal property. If, however, we choose $L=\{*\}$ and $p(*)=a_3$, then picking $T=\{*\}$ and $f(*)=a_1$ (or $a_2$), there is no $\bar f$ such that $p\circ\bar f = f$ since, for this $p$, $p\circ \bar f = p\neq f$.

Answer 2

The limit of a functor consists of both an object in the target category, as well as various arrows emanating from this object such that all the various triangles you get in the definition of the cone commute. So yes, the limit of $X$ will consist of a set $\lim_I X$, particularly a subset of $A\times B$, but it also carries the data of certain arrows $p_a\colon\lim_IX\to A$ and $p_b\colon\lim_I X\to B$, which are just the projection maps in the case of $\textsf{Set}$-valued functors.

One reason your singleton set isn't a limit in the example of your picture is because you can't factor the cone $(\lim_I X,p_a,p_b)$ through it. Let $\iota_A\colon\{\ast\}\to A$ be the chosen map from the singleton into $A$. Any factorization $\lambda\colon\lim_I X\to\{\ast\}$ would have to satisfy $p_a=\iota_A\circ\lambda$, for instance. But this is not possible since you can choose an element of $\lim_I X$ such that applying $p_a$ hits an element that $\iota_A$ does not.