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

Question

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.

Answer 1

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.