Timeline for What is an example of a colimit-dense generator which is not dense?
Current License: CC BY-SA 3.0
8 events
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| Nov 24, 2015 at 2:10 | comment | added | Omar Antolín-Camarena | 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. | |
| Nov 23, 2015 at 19:04 | comment | added | Mike Shulman | 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. | |
| Nov 22, 2015 at 18:44 | comment | added | Tim Campion | 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. | |
| Nov 22, 2015 at 18:38 | vote | accept | Omar Antolín-Camarena | ||
| Nov 22, 2015 at 18:38 | comment | added | Omar Antolín-Camarena | 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 $1< \mathrm{dim}\;V<\infty$. | |
| Nov 22, 2015 at 18:24 | comment | added | Tim Campion | 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. | |
| Nov 22, 2015 at 11:54 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
Fix accents on Adamek, provide URL to the book.
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| Nov 22, 2015 at 10:17 | history | answered | Jiří Rosický | CC BY-SA 3.0 |