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 $\gamma f'= k'\beta$?



Is this square commutative?
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. 


