# associated sequence is a short exact sequence iff the original sequence is a split short exact sequence.

Let $0 \to L \to M \to N \to 0$ be a sequence of $R$-modules. Then to prove that the associated sequence $$0 \to Hom_R(D, L) \to Hom_R(D, M) \to Hom_R(D, N) \to 0$$ is a short exact sequence of abelian groups for all $R$-modules $D$ if and only if the original sequence is a split short exact sequence.

• What's the question? Nov 4, 2017 at 6:39
• I have edited the question..@LordSharktheUnknown Nov 4, 2017 at 6:39
• Take $D=N{}{}$. Nov 4, 2017 at 6:40

Let $$0\rightarrow L\xrightarrow{\psi}M\xrightarrow{\phi}N\rightarrow 0$$ be the given sequence of $$R$$ modules.
We want to show that $$0\rightarrow \text{Hom}_R(D,L)\xrightarrow{\psi'}\text{Hom}_R(D,M)\xrightarrow{\phi'}\text{Hom}_R(D,N)\rightarrow 0$$ is exact for all $$R$$-modules $$D$$ if and only if $$0\rightarrow L\xrightarrow{\psi}M\xrightarrow{\phi}N\rightarrow 0$$ is a split short exact sequence.
($$\Rightarrow$$) Assume $$0\rightarrow \text{Hom}_R(D,L)\xrightarrow{\psi'}\text{Hom}_R(D,M)\xrightarrow{\phi'}\text{Hom}_R(D,N)\rightarrow 0$$ is exact for all $$D$$. Taking $$D=R$$ we have $$0\rightarrow \text{Hom}_R(R,L)\xrightarrow{\psi'}\text{Hom}_R(R,M)\xrightarrow{\phi'}\text{Hom}_R(R,N)\rightarrow 0$$ is exact. And since $$\text{Hom}_R(R,X)\cong X$$ for any $$R$$-module $$X$$, $$0\rightarrow L\xrightarrow{\psi}M\xrightarrow{\phi}N\rightarrow 0$$ is exact. To show that this sequence splits it suffices to find a splitting homomorphism $$\mu$$, (i.e. a homomorphism $$\mu:N\to M$$ such that $$\phi\circ\mu=\text{id}$$). Setting $$D=N$$, we have that $$0\rightarrow \text{Hom}_R(N,L)\xrightarrow{\psi'}\text{Hom}_R(N,M)\xrightarrow{\phi'}\text{Hom}_R(N,N)\rightarrow 0$$ is exact, in particular $$\phi':\text{Hom}_R(M,N)\rightarrow \text{Hom}_R(N,N)$$ is surjective. Therefore there must exist a $$\mu\in \text{Hom}_R(M,N)$$ such that $$\phi\circ\mu=\text{id}$$ (and such a $$\mu$$ is a splitting homomorphism).
($$\Leftarrow$$) Assume that $$0\rightarrow L\xrightarrow{\psi}M\xrightarrow{\phi}N\rightarrow 0$$ is a split exact sequence. From exactness it follows that $$0\rightarrow \text{Hom}_R(D,L)\xrightarrow{\psi'}\text{Hom}_R(D,M)\xrightarrow{\phi'}\text{Hom}_R(D,N)$$ is exact for all $$D$$, since Hom is a left exact functor. So to finish showing that $$0\rightarrow \text{Hom}_R(D,L)\xrightarrow{\psi'}\text{Hom}_R(D,M)\xrightarrow{\phi'}\text{Hom}_R(D,N)\rightarrow 0$$ is exact, we just have to show that $$\phi'$$ is surjective:
Let $$\rho\in\text{Hom}_R(D,N)$$. Since $$0\rightarrow L\xrightarrow{\psi}M\xrightarrow{\phi}N\rightarrow 0$$ splits, there exists a $$\mu:N\to M$$ such that $$\phi\circ\mu=\text{id}$$. In particular, we have $$\mu\circ\rho \in \text{Hom}_R(D,M)$$, so that $$\phi'\circ(\mu\circ\rho)=(\phi\circ\mu)\circ\rho=\text{id}\circ\rho=\rho,$$ confirming surjectivity.