Seeing $\mathbb{Q}$ as an ordered set, the colimit of a diagram $D:\mathcal{I} \to \mathbb{Q}$, when it exists, is just $\operatorname{colim}D \cong \operatorname{sup}_iD(i)$.

It seems to me that given any diagram $D:\mathbb{Q} \to \mathcal{E}$ into a cocomplete category $\mathcal{E}$, it extends to a functor $L:\mathbb{R}\cup\{-\infty,+\infty\} \to \mathcal{E}$ given by $L(r)=\operatorname{colim}\{D(q):q \leq r, q\in \mathbb{Q}\}\in \mathcal{E}$.

This establishes $\mathbb{R}\cup\{-\infty,+\infty\}$ as the free co-completion of $\mathbb{Q}$. Moreover, we obtain the density result asserting that every real number is the sup of all smaller rational numbers as a corollary of the fact that every presheaf is a colimit of representables.

Proof: $L$ is a functor preserving colimits and extending $D$ by construction. Given another $L'$ with these properties, since every $r$ is the colimit (= sup) of all its smaller rational numbers, it must be $L \cong L'$.

Am I stating something wrong here? I just thought about it, and it seems a nice and elementary example, but I have not seen it stated anywhere, which is very strange, and makes me suspect I am missing something and there's something wrong?

  • 1
    $\begingroup$ Perhaps this answer answers your question? $\endgroup$
    – varkor
    Commented Jun 9, 2023 at 15:49
  • $\begingroup$ @varkor not really. I mean the linked post in the question seems to say what I am saying, and the answer given does not convince me at all that it's wrong... $\endgroup$
    – aiaiai
    Commented Jun 9, 2023 at 15:54

2 Answers 2


The extension $L$ you describe does not preserve colimits in general. For instance, let $\mathcal{E}$ be the poset $\{0,1\}$ and let $D$ send the negative rationals to $0$ and the nonnegative rationals to $1$. Then your extension $L$ will send $[-\infty,0)$ to $0$ and $[0,\infty]$ to $1$. This does not preserve colimits because $\sup [-\infty,0)=0$.

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    $\begingroup$ Note this shows $[-\infty,\infty]$ isn't even the free cocompletion as a poset, let alone as a category. $\endgroup$ Commented Jun 9, 2023 at 17:31

The free cocompletion of any category is its category of presheaves. The category of presheaves on $\mathbb Q$ is the category of families of sets, contravariantly indexed by rational numbers. This is a lot bigger than $\mathbb R,$ in particular, it's not a small category (neither is any category of presheaves on a nonempty category.) The Yoneda embedding of $\mathbb Q$ into its presheaf category also fails to preserve coproducts of pairs: the coproduct of the presheaves represented by $1/3$ and $1/2$ has two elements in its values at any rational $x\le 1/3,$ so it's not equal to the presheaf represented by $1/2=1/3\vee 1/2.$ This behavior is impossible in any poset, which gives an example of a cocontinuous functor out of $\mathbb Q$ which cannot be cocontinuously extended to any poset containing $\mathbb Q.$

As discussed at the question Varkor linked in the comments, $[-\infty,\infty]$ is the Dedekind-Macneille completion of $\mathbb Q$ as a poset, a far more restrictive kind of cocompletion than the free cocompletion as a category. Intermediate is the free cocompletion of $\mathbb Q$ as a poset, its poset of downward-closed sets, which the linked question originally confused with $[-\infty,\infty].$


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