## A condition equivalent to splitting for a fixed short exact sequence

Let $R$ be a ring and let $0 \rightarrow L \stackrel{\varphi}{\rightarrow} M \stackrel{\psi}{\rightarrow} N \rightarrow 0$ be a short exact sequence of left, unital $R$-modules. (Denote this sequence by the pair $(\varphi,\psi)$.)

1. Prove that the associated sequence $0 \rightarrow \mathsf{Hom}_R(D,L) \stackrel{(\varphi \circ)}{\rightarrow} \mathsf{Hom}_R(D,M) \stackrel{(\psi \circ)}{\rightarrow} \mathsf{Hom}_R(D,N) \rightarrow 0$ (abbreviated $((\varphi \circ), (\psi \circ))$) is short exact for all $R$-modules $D$ if and only if $(\varphi,\psi)$ is split.
2. Prove that the associated sequence $0 \rightarrow \mathsf{Hom}_R(N,D) \stackrel{(\circ \psi)}{\rightarrow} \mathsf{Hom}_R(M,D) \stackrel{(\circ \varphi)}{\rightarrow} \mathsf{Hom}_R(L,D) \rightarrow 0$ (abbreviated $((\circ \psi),(\circ \varphi))$) is short exact for all $R$-modules $D$ if and only if $(\varphi,\psi)$ is split.

1. Suppose $(\varphi,\psi)$ is a split short exact sequence. That is, $\varphi$ is injective, $\psi$ is surjective, and $\mathsf{im}\ \varphi = \mathsf{ker}\ \psi$, and there exist module homomorphisms $\theta : N \rightarrow M$ such that $\psi \circ \theta = \mathsf{id}$ and $\lambda : M \rightarrow L$ such that $\lambda \circ \varphi = \mathsf{id}$. Now let $D$ be any left unital $R$-module, and define the sequence $((\varphi \circ),(\psi\circ))$ as above. We with to show that this sequence is short exact; it suffices to show that $(\psi \circ)$ is surjective. To that end, suppose $\alpha : D \rightarrow N$ is a module homomorphism. Note that $\theta \circ \alpha : D \rightarrow M$ is a module homomorphism, and moreover that $(\psi \circ)(\theta \circ \alpha) = \psi \circ \theta \circ \alpha$ $= \mathsf{id} \circ \alpha = \alpha$. So $(\psi \circ)$ is surjective, and the associated sequence $((\varphi \circ), (\psi \circ))$ is short exact for all $R$-modules $D$.

Conversely, suppose $((\varphi \circ),(\psi \circ))$ is short exact for all $D$. Then in particular, for $D = N$ the map $(\varphi \circ) : \mathsf{Hom}_R(M,N) \rightarrow \mathsf{Hom}_R(N,N)$ is surjective. Suppose $\varphi \circ \alpha = \mathsf{id}$; certainly then $\alpha$ is a splitting homomorphism for $(\varphi,\psi)$.

2. Suppose that $(\varphi,\psi)$ is a split short exact sequence. That is, $\varphi$ is injective, $\psi$ is surjective, $\mathsf{im}\ \varphi = \mathsf{ker}\ \psi$, and there exists homomorphisms $\theta : N \rightarrow M$ and $\lambda : M \rightarrow L$ such that $\psi \circ \theta = \mathsf{id}$ and $\lambda \circ \varphi = \mathsf{id}$. Now let $D$ be any left unital $R$-module and let $((\circ \psi),(\circ \varphi))$ be the sequence as defined above. Again, we wish to prove that this sequence is short exact; it suffices to show that $(\circ \varphi)$ is surjective. To that end, let $\alpha : L \rightarrow D$ be a module homomorphism. Note that $\alpha \circ \lambda : M \rightarrow D$ is a module homomorphism, and that $(\circ \varphi)(\alpha \circ \lambda) = \alpha \circ \lambda \circ \varphi$ $= \alpha \circ \mathsf{id} = \alpha$. Thus $(\circ \varphi)$ is surjective, and $((\circ \psi),(\circ \varphi))$ is short exact as desired for all modules $D$.

Conversely, suppose $((\circ \psi),(\circ \varphi))$ is short exact for all modules $D$. In particular, with $D = L$, the map $(\circ \varphi) : \mathsf{Hom}_R(M,L) \rightarrow \mathsf{Hom}_R(L,L)$ is surjective. Then there exists a homomorphism $\theta : L \rightarrow M$ such that $\theta \circ \varphi = \mathsf{id}$. This the sequence $(\varphi,\psi)$ splits.