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Nov 15, 2010 at 4:20 comment added Emerton Dear Anton, You're welcome.
Nov 14, 2010 at 15:29 comment added Emerton Yes, this is right.
Nov 14, 2010 at 9:46 comment added Buschi Sergio I think the condition is: $Ker(Fg)= Im(Ff)$ and no $Ker(Fg)= Cok(Ff)$ .
Nov 14, 2010 at 5:33 history answered Emerton CC BY-SA 2.5