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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 $f_{*}(\phi) = \phi \circ f$ for all $\phi \in Hom_A(M,D)$ and $g_{*} = \psi \circ g$ for all $\psi \in Hom_A(M'',D)$.

I am having trouble showing $\ker(f_{*}) \subset Im(g_{*})$. I will lists the steps I have taken up until where I got stuck:

Let $\phi \in \ker(f_{*}) \Rightarrow f_{*}( \phi) = 0 $ for all $ \phi \in Hom_A(M,D)$

$\Rightarrow \phi \circ f = 0 \Rightarrow Im(f) \subset \ker(\phi)$ By exactness then we have $\ker(g) \subset \ker(\phi)$ and this is where I am stuck.

How do we conclude from the last statement that there exists an induced map $\psi : M'' \rightarrow L$ such that $\psi \circ g = \phi$?

I have in my notes that $\psi = (\phi/\ker(g) ):M'' \rightarrow N$ but I dont really understand this notation. What is it about the containment of the kernel of g that induces the map?

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Looks like a duplicate of this question. (but maybe there's a better fit without dualization?) – t.b. Sep 7 '11 at 1:55
up vote 2 down vote accepted

Build a new morphism $\bar{\phi}:M''\longrightarrow D$ from $\phi$, defined by $\bar{\phi}(y)=\phi(x)$ where $y=g(x)$, then $g_*(\bar{\phi})=\phi$. First one should prove that the definition of $\bar{\phi}$ doesn't depend of the representant, that is if $g(x)=y=g(z)$ then $\phi(x)=\phi(z)$.But $x-z\in\ker g$, but as you nota, $\ker g\subseteq \ker \phi$, then $\phi(x-z)=0$ that implies $\phi(x)=\phi(z)$. At last $g_*(\bar{\phi})=\bar{\phi}g$, so for an $x\in M$ we got $\bar{\phi}(g(x))=\phi(x)$, so $\bar{\phi}g=\phi$.

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