Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
2  
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
add comment

1 Answer

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?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.