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.
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.
you can find a lot of examples of $Hom(_,G) failing to be exact even on this page. http://www-users.math.umn.edu/~garrett/m/repns/notes_2014-15/04b_adjoints_exactness.pdf