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Let $A$ be a commutative ring with identity. Let $D,M,M',M''$ be $A$-modules and suppose that $0\rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence. Label the maps $f:M'\rightarrow M$ and $g: M\rightarrow M''$.

Consider the induced sequence $0 \rightarrow Hom_A(M'',D) \rightarrow Hom_A(M,D) \rightarrow Hom_A(M',D)$ and label the map $f_{*} : Hom_A(M,D) \rightarrow Hom_A(M',D)$ given by $f_{*}(\phi) = \phi(f)$ for all $\phi \in Hom_A(M,D)$. I know the induced sequence is exact but I am missing a step in the proof.

How do we prove that the map $f_{*}$ is surjective? Or does $f_{*}$ need not be surjective for the sequence to be exact since we do not have the map $Hom_A(M',D) \rightarrow 0$?

I don't know if this is the place in module theory where you use baers criterion but I am having trouble filling in this step.

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Should it not read $f_*(\phi)=\phi \circ f$ for all $\phi\in Hom_A(M,D)$? – Jyrki Lahtonen Sep 6 '11 at 18:16
You use Baer's criterion to show that a module $D$ is injective, which is to say that $\operatorname{Hom}(-, D)$ is exact. Also, you don't need $M' \to M$ to be an injection here. – Dylan Moreland Sep 6 '11 at 18:39
up vote 2 down vote accepted

Hint: Check what happens when $A=D=M=M'=\mathbf{Z}$, $M''=\mathbf{Z}_2$, $f$ is multiplication by $2$ and $g$ is the natural projection.

Thinking: Exactness of the first sequence means (among other things) that $M'$ may be viewed as a submodule of $M$. Then the mapping $f_*$ amounts to restricting the domain of $\phi$ from $M$ to the submodule $M'$. From this point of view surjectivity of $f_*$ means that any homomorphism defined on the submodule $M'$ is gotten as a restriction of a homomorphism defined on the bigger module $M$. Does that always happen?

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