Straightforward application of Exactness of sequence I am trying to show that if the sequence $$0 \rightarrow M^* \xrightarrow{f^*} MCyl(\alpha)^* \rightarrow MC(id_L^*) \rightarrow 0$$
is exact ($\alpha^* : L^* \rightarrow M^*$, where $L^*, M^*$ are cochain complexes), then $f$ is a quasi-isomorphism. 
$MCyl, MC$ are the mapping cylinder and mapping cone respectively.
I know that $f^*$ quasi-isomorphism is equivalent to $MC(f)^*$ being exact but $MC(f)^*$ has messy objects and showing exactness is even messier. Is there a less messy approach?
 A: There is a less messy approach. The mapping cone $MC(id_{L^*})$ of the identity is sometimes just called the "cone" of $L^*$, and denoted by $CL^*$. The cone has degree $n$ part $L^{n+1}\oplus L^n$, with differentials:
$$d^n:L^{n+1}\oplus L^n\to L^{n+2}\oplus L^{n+1},\quad d^n(x,y) = (-d_L^{n+1}(x),x+d_L^n(y)).$$
With this knowledge alone, you can easily show that $CL^*$ is exact and so $H^n(CL^*)=0$ for all $n$. Now, the short exact sequence
$$0\xrightarrow{\ \ \ }M^*\xrightarrow{\ f^* \ }MCyl(\alpha)^*\xrightarrow{\ \ \ }CL^*\xrightarrow{\ \ \ }0$$
gives rise to a long exact sequence of cohomology modules:
$$\cdots\xrightarrow{\ \ \ }H^{n-1}(CL^*)\xrightarrow{\ \ \ }H^n(M^*)\xrightarrow{\ f^* \ }H^n(MCyl(\alpha)^*)\xrightarrow{\ \ \ }H^n(CL^*)\xrightarrow{\ \ \ }\cdots$$
but since $H^k(CL^*)=0$ for all $k$, the above reduces to short exact sequences:
$$0\xrightarrow{\ \ \ }H^n(M^*)\xrightarrow{\ f^* \ }H^n(MCyl(\alpha)^*)\xrightarrow{\ \ \ }0.$$
Thus, $f^*$ is a quasi-isomorphism.
Fun fact: One can even show that $CL^*$ is split exact, and this is analogous to the notion of being a contractible topological space, via the ideas Najib Idrissi explains in this answer.
