# Splitness of commutative diagrams


$$\begin{array}{c} 0 & \ra{} & A & \ra{f} & B & \ra{g} & C & \ra{} & 0 \\ & & \da{\gamma} & & \da{\beta} & & \da{\alpha} \\ 0 & \ra{} & D & \ra{k} & E & \ra{h} & F & \ra{} & 0 \end{array}$$

Is there any sufficient and necessary conditions to deduce that there are maps $g':C\to B$ and $h':F\to E$ satisfying the following axioms?

• $gg'=1_C$ and $hh'=1_F$
• $\beta g'=h'\alpha$
• Is this too tautological? Let $A = \mathrm{Fun}(\cdot \to \cdot, R\text{-Mod})$. The exact sequence splits iff $\beta$ represents the zero element of $\mathrm{Ext}_A^1(\alpha,\gamma)$. – Dan Petersen Dec 6 '13 at 14:54

I don't know if this will help, but this is what I make of it. First, clearly necessary conditions will be that top and bottom rows split. Thus we may as well suppose that the top row is $$0\to C\mathop{\to}\limits^{\pmatrix {1\cr 0}} A\oplus C \mathop{\to}\limits^{\pmatrix{0&1}} C\to0$$ and similarly for the bottom row (I am not going to try to make that diagram). By commutativity, there is a map $\sigma:C\to D$ such that $\beta=\pmatrix{\gamma&\sigma\cr 0&\alpha}$. A splitting of $g$ must have matrix $\pmatrix{u\cr1}$ for some $u:C\to A$ and similarly a splitting of $h$ requires a map $v:F\to D$. Finally, commutativity requires that $\gamma u+\sigma=v\alpha$. The splitting of the rows together with the existence of $u$ and $v$ satisfying that equation is necessary and sufficient. Although the map $\sigma$ is not unique, however $\sigma$ is chosen these conditions will remain.