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?

$\endgroup$
  • $\begingroup$ Looks like a duplicate of this question. (but maybe there's a better fit without dualization?) $\endgroup$ – t.b. Sep 7 '11 at 1:55
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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