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Nov 23, 2012 at 18:50 vote accept Mikhail Bondarko
Nov 14, 2012 at 14:11 history edited Mikhail Bondarko CC BY-SA 3.0
edited body; edited title
Nov 14, 2012 at 10:00 history edited Mikhail Bondarko CC BY-SA 3.0
I replaced $Hom(X,X/2X)$ by $End(X/2X)$ in condition 1, and added an explanation of condition 2.
Nov 14, 2012 at 9:49 comment added Mikhail Bondarko Yes, you are right! I updated the question.
Nov 14, 2012 at 9:48 history edited Mikhail Bondarko CC BY-SA 3.0
In the previous version of my question condition 2 was easily seen to contradict condition 1. So I replaced $Hom(X,X/2X)$ by $End(X/2X)$.
Nov 14, 2012 at 9:43 comment added Fernando Muro Whatever your category is, $\hom(X,X/2X)$ is a right $\hom(X,X)$-module where the action of $2\cdot 1_X$ is trivial, hence $2\cdot\hom(X,X/2X)=0$.
Nov 14, 2012 at 8:58 answer added Neil Strickland timeline score: 3
Nov 14, 2012 at 6:14 history asked Mikhail Bondarko CC BY-SA 3.0