# Does exactness of a sequence of groups imply exactness of the dual?

If $$A \rightarrow B \rightarrow C$$ is exact, with morphisms a and b respectively, that is $$Im(a)=ker(b)$$, then $$C^*\rightarrow B^* \rightarrow A^*$$ is also exact?

$$A^*=Hom(A,G)$$

A, B, C and G are groups, but I guess it might be also a good question in the case of modules.

The inclusion $$Im(b^*)\subset ker(a^*)$$ is easy, but the other inclusion is quite difficult to prove (maybe not even true)

Any ideas?

Edit: I was able to prove the statement in the case b has a right inverse. Maybe this is a necessary condition, I will check it later.

Edit 2: The splitting condition is sufficient but not necessary. As was mentioned in the answers, one could choose G to be an injective object and the sequence would always be exact.

• You mean abelian groups? Commented Nov 17, 2020 at 14:44
• In general, the Hom functors are left exact but not exact. Commented Nov 17, 2020 at 14:47
• Could be, does it work for the abelian case? Commented Nov 17, 2020 at 14:47
• I just found that it might be true if I add the condition that b has a right inverse Commented Nov 17, 2020 at 14:48
• then it is clear as the sequence will split and hom(_,G) is additive, otherwise you need that G is injective, which is precisely one of the definitions of being an injective object. Commented Nov 17, 2020 at 15:07

In general not, you need that $$G$$ is injective, as that is precisely one of the definitions of an injective objects, for abelian groups this is equivalent to being divisible.