Let C and C' be two abelian categories and let F:C---> C' a (additive?) functor. Suppose that "X non zero implies F(X) non zero". Does this imply that F is faithfull?
2007-11-27 05:06:06 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
F need not be faithful (even if it's additive). consider the functor F:C ---> C given by F(X) = X, F(f) = f for all f in Hom(X,X) and F(g) = 0 for all g in Hom(X,Y) for X not equal to Y. this functor satisfies your condition that X nonzero implies F(X) nonzero trivially. also, F is additive since it induces the identity homomorphism on Hom(X,X) and the zero homomorphism on Hom(X,Y). so, provided that C has two distinct nonzero objects X and Y with at least one nonzero morphism in Hom(X,Y), we have an additive functor which is not faithful.
2007-11-28 04:44:27 · answer #1 · answered by lkjh 3 · 1⤊ 0⤋
The functor C â D is one which maps every object of C to a fixed object X in D and every morphism in C to the identity morphism on X. Such a functor is called a constant or selection functor.
Maybe you can take C=C' and the functor to be the constant functor.
For example C=Ab, the category of abelian groups and
the functor that sends a group to say, (Z,+).
That would show that it's not necessarily to have a faithfull functor. The forgetfull functors are faithfull though.
2007-11-28 09:47:20 · answer #2 · answered by Theta40 7 · 1⤊ 0⤋
www. ask.com
2007-11-27 13:09:24 · answer #3 · answered by Anonymous · 0⤊ 3⤋
no sorry
2007-11-27 13:08:00 · answer #4 · answered by Sara C 2 · 0⤊ 3⤋