Exactness of a sequence in Abelian category

Question

Suppose we have a sequence in an Abelian Category, which consists of two morphism,lets say $f$:X-->Y, and a morphism $g$:Y-->Z such that gof=0 .If for every object M of the category the functor Hom(M,-) gives us an exact sequence, is it possible that the initial one could not be exact??

Answer 1

No, and actually the initial sequence will be "more than exact".

Indeed, take $M= \ker(g)$, then we have the following exact sequence $\hom(\ker(g), X)\to \hom(\ker(g),Y)\to \hom(\ker(g),Z)$; and if you take the inclusion $i: \ker(g)\to Y$ in the middle, it obviously gets sent to $0$ in $\hom(\ker(g),Z)$, so that by exactness it can be written as $f\circ j, j :\ker(g)\to X$.

It follows that the sequence is exact, but also "split", in the sense that there is a splitting $\mathrm{im}(f)\to X$

So any such sequence is exact, but there are exact sequences such that this doesn't hold (e.g. $\mathbb{ Z\to Z/2\to 0}$, in fact sequences of the type $A\to B\to 0$ where there is no section $B\to A$ are "universal" examples in a sense)