Consider the attached commutative diagram of $R$-modules, where the central column is a complex, and every row is exact. Prove that the left and right columns are also complexes. Also prove that if any two of the columns are exact, so is the third.
I'm self working on problems on Aluffi's Algebra book. This is the very last problem on Chapter III. Any help in solving this much appreciated.
Lets label the arrows to make life easier. We have: \begin{align} &\partial^L_n\colon L_{n}\longrightarrow L_{n-1},\\ &\partial^M_n\colon M_{n}\longrightarrow M_{n-1},\\ &\partial^N_n\colon N_{n}\longrightarrow N_{n-1},\\ &\alpha_n \colon L_n \to M_n\\ &\beta_n \colon M_n \to N_n \end{align}
Now we want to show that $\partial^L_{i} \circ \partial^L_{i+1} = \partial^N_{i} \circ \partial^N_{i+1} = 0$. By commutativity of your diagram we have: \begin{equation} \alpha_{i-1}\circ\partial^L_{i} \circ \partial^L_{i+1} = \underbrace{\partial^M_{i}\circ\partial^M_{i+1}}_{\text{$=\ 0$}}\circ\alpha_{i+1} = 0 \end{equation} where $\partial^M_{i}\circ\partial^M_{i+1} = 0$ since the middle column is a complex. Hence $\partial^L_{i} \circ \partial^L_{i+1} = 0$, since $\alpha$ is injective.
This shows that the left-hand column is a complex. Can you do the same for the right?