Mapping cone in the homotopy category I am just reading this book about triangulated category. There is something I am not understanding here when author is defining $\alpha(f) : Y \rightarrow M(f)$ he is defining it as $\alpha(f)_n := (0,id_{Y_n})$ Similarly for $\beta$. What I am not understanding is that why can we define it like this ? $X \bigoplus Y$ is the categorical direct sum of X and Y, it is not like the regular cartesian product? Also, why would $\beta$ be a morphism, because the differential $X[1]$ carries a sign?
I also don't understand why we have a short exact sequence ?

 A: In abelian categories, finite direct sums and direct products are equivalent (even naturally isomorphic), so there is no harm in using the "coordinate" description of a morphism in to $M(f)_n = X_{n-1}\oplus Y_n$.
The remark about the sign (regarding $\beta$ and $X[1]$) is necessary for the following reason. Recall that $X[n]$ is the complex defined as $X[n]_k=X_{n+k}$ and $d_k^{X[n]}=(-1)^kd_{n+k}^X$. If one were to omit either the sign on the $d_{n-1}^X$ in the upper left corner of $d_{n}^{M(f)}$ or the sign in the definition of $d_{k}^{X[n]}$, the map $\beta$ would not be a morphism of complexes, since the required squares would not commute: the composition $d\circ\beta$ would not equal the composition $\beta\circ d$ (they would be off by a sign).
In detail: $d\circ \beta$ is the morphism $X_{n-1}\oplus Y_n \stackrel{(id_{X_{n-1}},0)}{\to} X_{n-1} \stackrel{-d_{n-1}^X}{\to} X_n$ which may seen to be the morphism $(-d_{n-1}^X,0)$ from $X_{n-1}\oplus Y_n\to X_n$. $\beta\circ d$ is the morphism $X_{n-1}\oplus Y_n \stackrel{d^{M(f)}_n}{\to} X_{n}\oplus Y_{n+1} \stackrel{(id_X,0)}{\to} X_n$ which may be seen to be the morphism $(-d_{n-1}^X,0)$ from $X_{n-1}\oplus Y_n\to X_n$. Dropping the sign on the differential for $X[1]$ would negate the first composite map, while dropping the sign in the definition of $d^{M(f)}$ would negate the second composite map. These computations may be made by "matrix multiplication" as long as one's careful of the order that things happen in and when one needs to consider a row vector instead of a column vector ($\alpha,\beta$ can't both be row or column vectors).
The fact that there is a short exact sequence is easy to verify. $\alpha$ is clearly a monomorphism (because the identity map is), $\beta$ is clearly an epimorphism (because the identity map is), and $\beta\circ\alpha=0$ is also clear.
