Projective modules in long exact sequences

Question

Let $A$ be a commutative ring (with unit), and let $(P_i)_i$ be projective $A$-modules sitting in a long exact sequence of $A$-modules:
$$0 \longrightarrow P_1 \stackrel{f_1}{\longrightarrow} P_2\stackrel{f_2}{\longrightarrow} \cdots \stackrel{f_{r-1}}\longrightarrow P_r \longrightarrow 0.$$
By definition I know that, independently for each $i$, I can find an $A$-module $P_i'$ such that $P_i\oplus P_i'$ is free. But can I find $A$-modules $(P_i')_i$ together with $A$-linear maps $(p_i:P_i'\to P_{i+1}')_i$ such that each $P_i\oplus P_i'$ is free and the sequence
$$0 \longrightarrow P_1\oplus P_1' \xrightarrow{f_1\oplus p_1}P_2\oplus P_2'\xrightarrow{f_2\oplus p_2}\cdots \xrightarrow{f_{r-1}\oplus p_{r-1}} P_r\oplus P_r' \longrightarrow 0 $$
is again a long exact sequence of $A$-modules?

Answer 1

Yes. This is easy. Since the original sequence is split. You can suppose it is a short exact sequence.$0\rightarrow P_1\xrightarrow {f_1}P_2\xrightarrow {f_2}P_3\rightarrow 0$. Suppose $P_1\coprod Q_1$ and $P_3\coprod Q_3$ are free modules. Then $0\rightarrow P_1\coprod Q_1\rightarrow P_2\coprod Q_1\coprod Q_3\rightarrow P_3\coprod Q_3\rightarrow 0$ with natural morphisms is a short exact sequence.The middle is free since it is split.