MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I with to consider the following statement:

If $C$ is a cocomplete category having a dense small full subcategory $D$, then $C$ is complete.

(a full subcategory $D$ is dense in $C$ if every element of $C$ is canonical colimit of elements of $D$...)

I think I know how to prove it (I give proof below), and I want someone to reassure me that this statement is true exactly as stated, as it seems a little bit surprising.

Proof sketch:

Consider the functor $Y : C \to psh(D)$, where $psh(D)$ denotes the category of presheaves on $D$ (Yoneda functor, i.e. $Y(X)(A) = Hom (A, X)$). $D$ being dense in $C$ is equivalent to this functor being fully faithful. Futhermore, we have a functor $F: psh(D) \to C$, namely, the one which extends the inclusion $D \to C$ by cocontinuity (as presheaf categories have the property of being free cocompletions: Then one can see that $Y$ is right adjoint to $F$. So this renders $C$ as reflective subcategory of $psh(D)$ ( Now, $psh(D)$ is complete, and so every reflective subcategory of it. hence, $C$ is complete.

Thank you, Sasha

share|cite|improve this question
Perhaps you should fill in the condition in your title. – Martin Brandenburg Nov 10 '10 at 18:36
"namely, the one which extends the inclusion $D \to C$ by cocontinuity." Can you explain that in more detail? – Martin Brandenburg Nov 10 '10 at 23:38
As Martin says, you should revise the title. Right now, it looks like you're asking "what condition should I add to cocompleteness to guarantee completeness?", but instead you're asking "Is it true that any cocomplete category with a dense small full subcategory is complete?" – Theo Johnson-Freyd Nov 11 '10 at 2:40
Thank you for your comments; I tried to fix the question according to them. – Sasha Nov 11 '10 at 9:21
up vote 5 down vote accepted

This is a well-known theorem, you can find it for example in Abstract and concrete categories - the Joy of Cats, Theorem 12.12. The proof there uses that a cocomplete category with a weakly terminal object has a terminal object (the preparation for the Freyd's adjoint functor theorem).

share|cite|improve this answer

As Martin says, this is true. I like to think of this implication as factoring through a third notion: any total category is complete (and more than complete), whereas any cocomplete category with a small dense subcategory is total.

share|cite|improve this answer
Well this is included in Sasha's proof, right? – Martin Brandenburg Nov 10 '10 at 23:40
Not exactly; it looks to me like he proved that the embedding of C into the presheaf category of its small dense subcategory D has a left adjoint, rather than that the embedding of C into its own presheaf category has a left adjoint. Close, though. – Mike Shulman Nov 11 '10 at 3:39
@Mike: Well, this is basically contained in his proof. You just have compose with the geometric morphism $(i_*,i^*),$ where $i:D \hookrightarrow C$ is the inclusion. Since $i$ is full and faithful, so is $i_*.$ – David Carchedi Jul 11 '12 at 20:37
Like I said: close. (-: – Mike Shulman Jul 11 '12 at 21:28
For future readers, I'd like to point out a reference to this fact: corollary 6.5, "A survey of totality for enriched and ordinary categories", by Max Kelly, where it is worked out in the enriched context. – Bruno Stonek Feb 6 at 7:43

A more direct proof:

A limit of a diagram $(A_j)_{j\in J}$ is the colimit of the following $\pi: \mathcal{D}_A\to \mathcal{C}$: Objects of $\mathcal{D}_A$ are the $(D, \pi_D)$ with $D\in \mathcal{D}$ with a (natural) cone $\pi_{D,j}: D \to A_j\ j\in J$, the morphisms $(D, \pi_D)\to (D', \pi_{D'})$ are morphims $D\to D'$ coherents by $\pi_D$ and $\pi_{D'}$ and $\pi: \mathcal{D}_A\to \mathcal{C}: (D, \pi_D) \mapsto D$ .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.