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Let $\mathscr{A}$ be an Abelian category, and let $$0\to A\xrightarrow{i}B\xrightarrow{q}C\to 0$$ be a short exact sequence in $\mathscr{A}$. Then it is a simple fact (proved by working with elements) that in the category of modules $\mathscr{A}=R\hbox{-}\mathbf{Mod}$, we have the following equivalence:

TFAE:

  1. $B=A\oplus C$, with inclusion maps $$A\xrightarrow{i}B\xleftarrow{j}C$$ and projection maps $$A\xleftarrow{p}B\xrightarrow{q}C$$ satisfying the usual identities.
  2. There exists some morphism $j:C\to B$ such that $q\circ j = \mathrm{id}_C$.
  3. There exists some morphism $p:B\to A$ such that $p\circ i = \mathrm{id}_A$.

Now it's trivial to show that 1) implies 2) and 3). If I can prove that 2) implies 1), than the case 3) implies 1) should probably follow by analogy. My current approach is to define the map $$B\xrightarrow{p}\mathrm{coker\,}j$$ and to somehow show that $p\circ i:A\to\mathrm{coker\,}j$ is an isomorphism. However, I can't seem to figure out how to do this.

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It's not so difficult to come up with a proof without elements, here's one. Let $q\colon B\twoheadrightarrow C$ be an epimorphism with a section $j\colon C\to B$ such that $$\tag{1} q\circ j = id_C.$$

  • As we have a short exact sequence $0\to A\xrightarrow{i} B\xrightarrow{q} C \to 0$, the morphism $i\colon A\rightarrowtail B$ satisfies the universal property of $\ker q$. In particular, we have $$\tag{2} q\circ i = 0.$$

  • We have $q\circ (id_B - j\circ q) = q−q\circ j\circ q = q - id_C \circ q = 0$, so by the universal property of kernels, there exists a canonical arrow $p\colon B \to A$ such that $i\circ p = id_B - j\circ q$; that is, $$\tag{3} i\circ p + j\circ q = id_B.$$

  • Note that $i \circ p \circ i = (id_B - j \circ q) \circ i = i - j ◦ \underbrace{q \circ i}_{= 0} = i$, and $i$ is a monomorphism, therefore $$\tag{4} p\circ i = id_{A}.$$

  • We also have $i\circ p\circ j = (id_B - j\circ q) \circ j = j - j\circ \underbrace{q\circ j}_{= id} = j - j = 0$, and $i$ is a monomorphism, therefore $$\tag{5} p\circ j = 0.$$

Now the identities (1)-(5) say that $B \cong A\oplus C$.

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