This is a problem that I thought at first was obvious but that became less clear the more I thought about it. Assume we have a finitely generated algebra $A$ over a field $k$, and a short exact sequence of projective $A$-modules $l\_1:\quad0\rightarrow P\_1 \rightarrow P\_2 \rightarrow P\_3 \rightarrow 0$. Let's assume we can lift this short exact sequence to their generators, i.e., let $F\_i$ be a free module that surjects onto $P\_i$ and assume there is a short exact sequence $l\_2:\quad0\rightarrow F\_1 \rightarrow F\_2 \rightarrow F\_3 \rightarrow 0$ that surjects onto $l\_1$. Furthermore assume that all morphisms in $l\_2$ are given by matrices in $M\_{n\times m}(k)$ for suitable values of $m$ and $n$. This implies that we can choose splitting morphisms for $l\_2$ that are also in $M\_{n\times m}(k)$. My question: is there a such a choice of a splitting for $l\_2$ that induces a splitting on $l\_1$? This is equivalent to asking whether such a splitting will map the kernels of $F\_i\rightarrow P\_i$ to each other. (If the answer depends on $A$, for what $A$'s is it “yes”?)



When splitting of short exact sequence preserves the kernels
The answer is "no" unless $A=k$.


Let $a\in A\setminus k$, and let $l\_2$ and $l\_1$ be the first and second rows of the commutative diagram
$$\require{AMScd}
\begin{CD}
0@>>>A@>\begin{pmatrix}1\\0\end{pmatrix}>>A^2
@>\begin{pmatrix}0&1\end{pmatrix}>>A@>>>0\\
@.@VV1V@VV\begin{pmatrix}1&a\end{pmatrix}V@VVV\\
0@>>>A@>1>>A@>>>0@>>>0\\
@.@VVV@VVV@VVV\\
@.0@.0@.0@.
\end{CD}
$$


