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$\newcommand{\colim}{\operatorname{colim}} \newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Elt}{\operatorname{Elt}} \newcommand{\Func}{\operatorname{Func}}$ Let $C$ be a small category with finite limits, and let $D$ be a category with (small) colimits and finite limits. If $G : C \rightarrow D$ preserves finite limits, then is $\colim G$ a terminal object of $D$?

Background

I'm trying to understand the equivalence of categories $\Func_{lex}(C,D) \simeq \Func_{lexcc}(\Psh(C),D)$. The map from right to left is simple (just pre-composition with the Yoneda embedding), but going from left to right is more tricky. Given $G : C \rightarrow D$, we form $F_G : \Psh(C) \rightarrow D$ as follows. If $E \in \Psh(C)$, we can form the category of elements $\Elt E$, which comes with a functor $\pi : \Elt E \rightarrow C$, and then we let $F_G(E) = \colim(G \circ \pi)$.

I'm trying to understand why this latter map preserves lex functors, and so I'm starting with the simplest case: preserving the terminal object. So, assume $G$ preserves the terminal object; can we show $F_G$ does too? The terminal object $T \in \Psh(C)$ is computed pointwise, and it's easy to show that $\pi : \Elt T \rightarrow C$ is an isomorphism, so $F_G(T) = \colim G$. Thus the claim reduces to checking that $\colim G$ is a terminal object of $D$.

Why am I posting this self-answered question?

  1. When I first came upon this problem, I couldn't find any answers online, and I suspected it to be false (it was too simple for me to not know). To my surprise, it turned out to be true, so I wanted to make this post to help others find an answer more easily.
  2. I'm hoping that others may add answers that provide a more high-level view of why this is true, instead of just the direct calculation. (And indeed, the link in my answer, which I only discovered when writing this question, already provides a generalization.)
  3. My answer links to another question whose answer provides a solution to mine. While this might suggest a duplicate, I believe that the angle of my question is different enough that it is worth preserving. That is, while the same underlying phenomenon answers both questions, the questions themselves are different enough that searching for one may not turn up the other (indeed, this was the case for me).
Sambo
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1 Answers1

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$\newcommand{\colim}{\operatorname{colim}}$ The answer is yes, and indeed, the situation is more general. We have the following fact:

Let $C,D$ be categories with $C$ small. If $C,D$ have terminal objects, and $G : C \rightarrow D$ preserves the terminal object, then $\colim G$ exists and is a terminal object.

It is pretty easy to show directly that the terminal object $1 \in D$ satisfies the universal property of being the colimit of $G$ (it forms a cocone along with the unique morphisms $G(c) \rightarrow 1$ for each $c \in C$). However, we can also use the result from this other MSE answer: if $C$ has a terminal object $1$, then $\colim G \cong G(1)$. Then, adding the fact that $G$ preserves the terminal object immediately shows that $\colim G$ is terminal. (Note that the proof in that answer can be easily adapted to show directly that $1 \cong \colim G$.)

Sambo
  • 7,469