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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.

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    $\begingroup$ You mean abelian groups? $\endgroup$
    – Paul Frost
    Commented Nov 17, 2020 at 14:44
  • $\begingroup$ In general, the Hom functors are left exact but not exact. $\endgroup$
    – Hermetically Sealed Halibut
    Commented Nov 17, 2020 at 14:47
  • $\begingroup$ Could be, does it work for the abelian case? $\endgroup$
    – D18938394
    Commented Nov 17, 2020 at 14:47
  • $\begingroup$ I just found that it might be true if I add the condition that b has a right inverse $\endgroup$
    – D18938394
    Commented Nov 17, 2020 at 14:48
  • $\begingroup$ 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. $\endgroup$
    – Felix
    Commented Nov 17, 2020 at 15:07

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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

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  • $\begingroup$ thank you very much! I wasn't aware of this definition of an injective object. As you mentioned it's exactly my problem here, really interesting, thank you for the reference. $\endgroup$
    – D18938394
    Commented Nov 17, 2020 at 17:29
  • $\begingroup$ So the splitting condition isn't necessary, it's only sufficient. $\endgroup$
    – D18938394
    Commented Nov 17, 2020 at 17:31
  • $\begingroup$ exactly, if it splits however, it becomes a split ses and so it has to split in the image cat as well, as an additive functor sends direct sums to direct sums $\endgroup$
    – Felix
    Commented Nov 17, 2020 at 17:37

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