Timeline for Separators in the Category of Groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2015 at 8:00 | comment | added | Tom Church | @PyRulez: a group that maps nontrivially to $\mathbb{Z}$ surjects to $\mathbb{Z}$ (take the image of a nontrivial map). | |
| Jan 16, 2015 at 23:08 | comment | added | Christopher King | Wait, if one takes $G = \mathbb{Z}$ and then you have $s : G \rightarrow \mathbb{Z} = x \mapsto 3*x$ Is a separator for you $f$ and $g$ but no an epimorphism I think. I am probably wrong. If so, how? | |
| Jan 16, 2015 at 22:53 | comment | added | Christopher King | "This is a huge class of groups which has no particularly nice description beyond the definition, as far as I know." If you don't need the description to be nice, just use the yoneda lemma. | |
| Feb 14, 2012 at 1:54 | comment | added | Will Sawin | "Every separator has a map to it that is an epimorphism" ia a property but is not universal. It seems the most promising way to proceed is consider the set of groups and assignments of morphisms $x$ to pairs $f,g$, satisfying some kind of naturality/commutativity/coherence condition, such that these correspond to the category of "groups with a surjection onto $\mathbb Z$", of which $\mathbb Z$ is of course a final element. But I can't think of a way to do this right now. | |
| Feb 14, 2012 at 0:04 | history | edited | David White | CC BY-SA 3.0 |
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| Nov 2, 2009 at 23:09 | comment | added | Steven Gubkin | Thanks! I was not thinking. Do you have any insight on the second question? | |
| Nov 2, 2009 at 23:05 | vote | accept | Steven Gubkin | ||
| Nov 2, 2009 at 23:03 | history | answered | Tom Church | CC BY-SA 2.5 |