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Suppose that the following commutative diagram of $R$-modules is given. $ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$

$$ \begin{array}{c} A & \ra{f} & B & \ra{} & 0 \\ \da{\gamma} & & \da{\beta} & \\ D & \ra{k} & E & \ra{} & 0 \end{array} $$ Suppose that $f$ and $k$ are both split epimorphisms and $\gamma$ and $\beta$ are both split monomorphisms.

We know that $\beta f= k\gamma$. By splitness we know that there maps $f':B\to A$ and $k':E\to D$ such that $ff'=1_B$ and $kk'=1_E$.

Can we conclude that $\gamma f'= k'\beta$?

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No, take $A = D$, $B = E$, $\gamma$ the identity of $A$, $\beta$ the identity of $B$ and $f = k$ a split epimorphism from $A$ to $B$. Then the commutation you want is equivalent to the uniqueness of sections of $f$, which is wrong.

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    $\begingroup$ As @JeremyDaniel points out, not every choice of $f'$ and $k'$ works. But there is always a choice that does work. If you'd given a bit more context for your question, I might have a better idea of whether that fact is of interest to you. $\endgroup$
    – Jeremy Rickard
    Commented Dec 5, 2013 at 15:04

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