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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

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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).

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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.

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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. – lentic catachresis 1 hour ago

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 $\pi: \mathcal{D}_A\to \mathcal{C}: (D, \pi_D) \mapsto D$ .

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