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.

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