If $0 \rightarrow A_{\bullet} \rightarrow B_{\bullet} \rightarrow C_{\bullet} \rightarrow 0$ is a short exact sequence of chain complexes (of R-modules), then, whenever two of the three complexes $A_{\bullet}$,$B_{\bullet}$,$C_{\bullet}$ are exact, so is the third.
This exercise 1.3.1 in Weibel's Introduction to Homological Algebra.
I was trying to tackle the exercise via diagram chasing in the diagram
However, in all three situations I eventually get stuck. I am starting to think that diagram chase might not be the right approach here?
Thank you.
Best Answer
For completeness, I think I can cover the case where $A_{\bullet}$ and $B_{\bullet}$ are exact myself now. I will try to stick to the notations used in the other posts.
If $c \in C_n$ with $d(c)=0$, there is, by the surjectivity of $\beta_n$, a $b$ in $B_n$ such that $$\beta_n(b)=c.$$
Then $$\beta_{n-1}(d (b)) = d(\beta_n (b)) = d(c) = 0$$ and thus, by exactness of the n-1-th row, there is an $a$ in $A_{n-1}$ with $\alpha_{n-1}(a) = d(b)$.
For $a$ we have $$\alpha_{n-2}(d(a)) = d(\alpha_{n-1}(a)) = d(d(b)) = 0$$ and, since $\alpha_{n-2}$ is injective, $d(a)=0$ follows.
Then, by exactness of $A_{\bullet}$ we have an $a'$ in $A_n$, such that $d(a')=a$.
Consider $b-\alpha_n(a')$ in $B_n$. We have $$d(b - \alpha_n(a'))=d(b) - d(\alpha_n(a')) = d(b) - \alpha_{n-1}(d(a')) = d(b) - d(b) = 0,$$ thus by exactness of $B_{\bullet}$, there is a $b'$ in $B_{n+1}$, such that $d(b')=b-\alpha_n(a')$.
Finally, $\beta_{n+1}(b')$ is the desired pre-image of $c$ in $C_{n}$, since $$d(\beta_{n+1}(b'))=\beta_n(d(b'))=\beta_n(b) - \beta_n(\alpha_n(a')) = c.$$
Thanks, everyone!