Timeline for A question in category theory
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2013 at 12:52 | comment | added | Jeremy Rickard | For the "non-natural" case: Every $A$-module, for $A=k[x]/(x^2)$ ($k$ a field), is a direct sum of copies of $k$ and $A$. Taking $X$ to be a countable direct sum of copies of $k$ and $Y$ a countable direct sum of copies of $A$ gives a counterexample in the category of $A$-modules. Though this feels a bit like cheating. If $C$ is the category of finitely-generated modules for a finite-dimensional algebra (or, more generally, an Artin algebra), then there is no counterexample. I think this is due to Auslander, and in any case is an easy consequence of the existence of almost split sequences. | |
| May 9, 2013 at 9:17 | comment | added | Toink | I would be interested whether there exists a counterexample if the isomorphism is not assumed natural. if C doesn't have to be abelian it is easy: Take C to be the category freely generated by two objects and two arrows between them (in different directions). Then for any $X$,$Y$ (possibly $X=Y$) in C we have $Hom(X,Y)\cong \aleph_0$, but the two objects in C are not isomorphic. | |
| May 9, 2013 at 7:46 | comment | added | David Roberts♦ | If you're dealing with a non-locally small abelian category, then your problems are bigger than deciding whether two objects are isomorphic :-) | |
| May 9, 2013 at 6:54 | vote | accept | Stef | ||
| May 9, 2013 at 6:53 | comment | added | Stef | Yes, they are naturals and nowI realize that you are right, $Nat(h^X, h^Y)$ is in bijection with $ Hom_C(X, Y)$. Since this bijection behaves well with composition then it follows that to my natural transformation should correspond an isomorphism. Therefore I should impose only the conditions necessary to be satisfied Yoneda' s lemma, i.e the hom spaces to be sets. Thank you very much! | |
| May 9, 2013 at 6:45 | answer | added | jmc | timeline score: 4 | |
| May 9, 2013 at 6:38 | comment | added | David Roberts♦ | If the isomorphisms are natural, then you can use Yoneda to get $X\simeq Y$. | |
| May 9, 2013 at 6:34 | history | asked | Stef | CC BY-SA 3.0 |