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An object $G$ of a category $\mathcal{C}$ is a dense generator if every object $X$ is the colimit of the canonical diagram of copies of $G$ mapping to $X$. (This canonical diagram is indexed by the full subcategory of the slice $\mathcal{C}_{/X}$ on the objects of the form $G \to X$.)

An object $G$ is called a colimit-dense generator if every object $X$ is a colimit of some diagram of copies of $G$. (That means $X$ is the colimit of some functor $F : \mathcal{I} \to \mathcal{C}$ with $F(i) = G$ for all objects $i$.)

Is there an example of a colimit-dense generator which is not a dense generator?

Omar Antolín-Camarena
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    To forestall someone pointing this out, Mike Shulman in http://home.sandiego.edu/~shulman/papers/generators.pdf claims that $\mathbb{Z} \in \text{Ab}$ is colimit-dense but not dense. In fact it is not even colimit-dense! See http://mathoverflow.net/questions/204792/is-every-abelian-group-a-colimit-of-copies-of-z for a discussion. – Qiaochu Yuan Nov 22 '15 at 05:37

1 Answers1

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$\Bbb R$ is colimit-dense in the category of real vector spaces but not dense (see 6.F, 6.34 in my book with J. Adámek "Locally presentable and accessible categories").

Andrej Bauer
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Jiří Rosický
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    I like thinking of this in terms of the definition of dense generator which says that $G$ is dense in $C$ if the nerve functor $X \mapsto Hom(i_G-, X)$ is fully faithful (where $i_G: G \to C$ is the inclusion functor). It's easy to see that a natural transformation $Hom(i_{{\mathbb R}}, X) \implies Hom(i_{{\mathbb R}}, Y)$ consists of a homogeneous function $X \to Y$ but there's no reason for the function to be linear, since the addition map $\mathbb R^2 \to \mathbb R$ is not represented in the category ${\mathbb R}$. $\mathbb R^2$ is dense, though, because it does represent this map. – Tim Campion Nov 22 '15 at 18:24
  • Thanks, Professor Rosický! This is a very nice example. The canonical colimit of $\mathbb{R}$'s for a vector space $V$ gives you a vector space with basis the projectivization $\mathbb{P}(V)$ (another manifestation of what @Tim says, that $\mathbb{R}$ only imposes the correct restriction on scalar multiplication, not on addition), which is of greater dimension than $V$ for $1< \mathrm{dim};V<\infty$. – Omar Antolín-Camarena Nov 22 '15 at 18:38
  • It also suggests a generalization: in any variety where every algebra is free and there is at least one nontrivial operation of arity $\geq 2$, the free algebra on one generator fits the bill. These varieties have been classified: they are either modules over a division ring or affine spaces over a division ring. – Tim Campion Nov 22 '15 at 18:44
  • Excellent, now I can finally fix the note of mine that Qiaochu mentioned in a comment by including a correct example. This also suggests that perhaps my incorrect example was supposed to be $\mathbb{Z}$ in the category of free abelian groups. – Mike Shulman Nov 23 '15 at 19:04
  • In free abelian groups the $\mathbb{Z}$ example does seem to work, @MikeShulman. By the way, thanks a lot for writing that note and several other extremely useful things that you've written that aren't meant to be published (I would guess), such as blog posts. – Omar Antolín-Camarena Nov 24 '15 at 02:10