Vtyjkyuk

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.

Let $0rightarrow LxrightarrowpsiMxrightarrowphiNrightarrow 0$ be the given sequence of $R$ modules.

We want to show that
$$0rightarrow textHom_R(D,L)xrightarrowpsi'textHom_R(D,M)xrightarrowphi'textHom_R(D,N)rightarrow 0 $$
is exact for all $R$-modules $D$ if and only if $0rightarrow LxrightarrowpsiMxrightarrowphiNrightarrow 0$ is a split short exact sequence.

($Rightarrow$) Assume
$$0rightarrow textHom_R(D,L)xrightarrowpsi'textHom_R(D,M)xrightarrowphi'textHom_R(D,N)rightarrow 0 $$
is exact for all $D$. Taking $D=R$ we have
$$0rightarrow textHom_R(R,L)xrightarrowpsi'textHom_R(R,M)xrightarrowphi'textHom_R(R,N)rightarrow 0 $$
is exact. And since $textHom_R(R,X)cong X$ for any $R$-module $X$,
$$0rightarrow LxrightarrowpsiMxrightarrowphiNrightarrow 0$$
is exact. To show that this sequence splits it suffices to find a splitting homomorphism $mu$, (i.e. a homomorphism $mu:Nto M$ such that $phicircmu=textid$). Setting $D=N$, we have that
$$0rightarrow textHom_R(N,L)xrightarrowpsi'textHom_R(N,M)xrightarrowphi'textHom_R(N,N)rightarrow 0 $$
is exact, in particular $phi':textHom_R(M,N)rightarrow textHom_R(N,N)$ is surjective. Therefore there must exist a $muin textHom_R(M,N)$ such that $phicircmu=textid$ (and such a $mu$ is a splitting homomorphism).

($Leftarrow$) Assume that $0rightarrow LxrightarrowpsiMxrightarrowphiNrightarrow 0$ is a split exact sequence. From exactness it followings that
$$0rightarrow textHom_R(D,L)xrightarrowpsi'textHom_R(D,M)xrightarrowphi'textHom_R(D,M)$$
is exact for all $D$, since Hom is a left exact functor. So to finish showwing that
$$0rightarrow textHom_R(D,L)xrightarrowpsi'textHom_R(D,M)xrightarrowphi'textHom_R(D,M)rightarrow 0$$
is exact, we just have to show that $phi'$ is surjective:

Let $rhointextHom_R(D,N)$. Since $0rightarrow LxrightarrowpsiMxrightarrowphiNrightarrow 0$ splits, there exists a $mu:Nto M$ such that $phicircmu=textid$. In particular, we have $mucircrho in textHom_R(D,M)$, so that
$$phi'circ(mucircrho)=(phicircmu)circrho=textidcircrho=rho,$$
confirming surjectivity.