3
$\begingroup$

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?

  • 0
    Looks like a duplicate of [this question](http://math.stackexchange.com/questions/47401/hom-is-a-left-exact-functor). (but maybe there's a better fit without dualization?)2011-09-07

1 Answers 1

3

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 $\bar{\phi}g=\phi$.