$0\to L\to M\to N\to 0 $ is split if $0\to {\rm Hom}_R(D,L)\to {\rm Hom}_R(D,M)\to {\rm Hom}_R(D,N)\to 0$ is exact for any $D$.

Question

Prove that $0\rightarrow L\rightarrow M\stackrel{\phi}\rightarrow N\rightarrow 0 $ is split if $$0\rightarrow {\rm Hom}_R(D,L)\rightarrow {\rm Hom}_R(D,M)\rightarrow {\rm Hom}_R(D,N)\rightarrow 0$$ is exact for any $D$, and $R$ is a ring with $1$.

Proof : Hint is $D=N$. I have no idea.

[add] identity $i : D\rightarrow D$ has a lift $I : N\rightarrow M,\ \phi\circ I=id $

Note that ${\rm ker}\ \phi,\ I(N)$ are $R$-submodules and we have $$ {\rm ker}\ \phi \oplus I(N) \subset M$$

If $x\in M$ with $\phi (x)\neq 0$ then we have a claim $x_0\in I(N),\ x-x_0\in {\rm ker}\ \phi$

Answer 1

Hint: If we let $D=N$, there is a special homomorphism in $\mathrm{Hom}_R(N,N)$, namely the identity morphism on $N$. Since the sequence of Hom sets is exact, there must be some $R$-module map $N\to M$ which maps to the identity morphism in $\mathrm{Hom}_R(N,N)$. What does this mean with respect to the map $M\to N$ in the original sequence?