In a 3x3 diagram of modules, if the first two rows are split exact, is the third row also split? I have the following commutative diagram of $R$-modules
$$\begin{matrix}
0 & \to & L' & \to & L & \to & L'' & \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
0 & \to & M' & \to & M & \to & M'' & \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
 &  & N' & \to & N & \to & N''& \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
&& 0 && 0 && 0 &&\\
\end{matrix}$$
with exact rows and columns, where the first and second rows are split exact.
Question: Is it also true that the third row is split exact?
I have determined it is true if the splittings are compatible, i.e. the reverse square
$$\begin{matrix}
L & \leftarrow & L''\\
\downarrow && \downarrow\\
M & \leftarrow & M''\\
\end{matrix}$$
is commutative.
However, I'm not sure if that is the case in my situation.
Question 2: If $L'', M''$ are projective, can we choose a splitting of the second row which descends to a splitting of the third row?
Question 3: If the upper left square is Cartesian, and the first two rows split, will the third row split?
I have thought about this for a while, but I can't seem to make the argument work. I would appreciate any help.
 A: Generalizing this answer, every exact sequence $X\xrightarrow{\alpha} Y\xrightarrow{\beta} Z\to 0$ appears as the bottom row of a diagram of the kind you describe, since you can take the diagram
$$\require{AMScd}\begin{CD}
0@>>>0@>>>X@>1>>X@>>>0\\
@.@VVV@VV\begin{pmatrix}1\\\alpha\end{pmatrix}V@VV\alpha V\\
0@>>>X@>\begin{pmatrix}1\\0\end{pmatrix}>>X\oplus Y@>\begin{pmatrix}0&1\end{pmatrix}>>Y@>>>0\\
@.@VV1V@VV\begin{pmatrix}\alpha&-1\end{pmatrix}V@VV-\beta V\\
@.X@>\alpha>>Y@>\beta>>Z@>>>0\\
@.@VVV@VVV@VVV\\
@.0@.0@.0
\end{CD}$$
So no, the third row of your diagram does not have to be split exact: it can be absolutely anything.
Special cases of this also give negative answers to your other questions:
Taking $X\to Y\to Z\to0$ to be a projective presentation of a nonprojective module answers Question 2.
Taking $0\to X\to Y\to Z\to 0$ to be a nonsplit short exact sequence gives an example where the top left square is Cartesian, answering Question 3.
A: I think for your first question, that square always commutes. Since $L$ is a coproduct in an abelian category, we can use its universal property. The map $L' \to L$ in your diagram and the section map $L'' \to L$ which exists by splitting are the canonical inclusions, and we also have maps $L' \to M' \to M$ and $L'' \to M'' \to M$, so by the universal property, there is a unique map $L \to M$ making the coproduct diagram commute. But the diagram that you get for the universal property of the coproduct contains the top left square of the big diagram in the OP, and this map is unique, so the map appearing in the coproduct diagram must be the same as the map in OP's. Now we just note that the commutative square in Question 1 is the other half of the coproduct diagram, so we are done.
