TITLE: Intuition about the triangulation of a homotopy category K(A)
QUESTION [14 upvotes]: Let $\cal{A}$ be an additive category. Given a morphism of (cochain) complexes $f:X\rightarrow Y$ we can form the mapping cone $C_f$, which is the complex $X[1]\oplus Y$ with differential given by
 $$\partial^n=\begin{pmatrix} -\partial_X^{n+1} & 0 \\\ f^n & \partial_Y^n\end{pmatrix}.$$
We have a canonical sequence of complexes 
$$ C_f[-1]\overset{\tilde{\beta}}{\longrightarrow} X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$
where $\alpha$ is injection, $\beta$ is projection, and $\tilde{\beta}$ is just $\beta[-1]$.
One can show that, up to homotopy, the morphisms $\alpha$ and $\beta$ only depend on the homotopy class of $f$ and so the above situation descends to the case of a morphism $f:X\rightarrow Y$ in the homotopy category of complexes $K(\cal{A})$. Triangles
$$ X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$
in $K(\cal{A})$ obtained in this way are called "standard triangles". We obtain the structure of a triangulated category on $K(\cal{A})$ by taking our distinguished triangles to be all triangles which are isomorphic (in $K(\cal{A}))$ to standard triangles.
I am trying to develop intuition and understand the motivation behind this triangulation on $K(\cal{A})$. I imagine that in some sense these triangles should capture the homotopy-theoretic information of morphisms of complexes, but at this point I do not have a clear understanding of the significance of these triangles.
I do have some understanding about how (in some vague sense) the sequence
$$ C_f[-1]\overset{\tilde{\beta}}{\longrightarrow} X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$
captures homotopy-theoretic information of the morphism $f:X\rightarrow Y$. For example, the map $\alpha:Y\rightarrow C_f$ can be thought of as a kind of "homotopy cokernel" of $f$ and the map $\tilde{\beta}:C_f[-1]\rightarrow X$ can be regarded as a kind of "homotopy kernel" of $f$. More precisely, a morphism $g:Y\rightarrow Z$ factors through $\alpha$ iff the composition $g\circ f$ is homotopic to 0. Indeed, you can show that there is a bijection between factorizations of $g$ through $\alpha$ and the homotopies from $g\circ f$ to $0$. In particular, there is a canonical homotopy of $\alpha \circ f$ to $0$. There is an analogous situation for $\tilde{\beta}$. One consequence of the above is that two morphisms of complexes $f_1,f_2:X\rightarrow Y$ are homotopic iff $f-g$ factors through the canonical map $X\rightarrow C_{id_X}$.
Because of such results I have some vague idea of how this business with the mapping cone of a morphism is well-entangled with questions of homotopy, but I would like a more precise understanding of the sense in which 

the standard triangle $ X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $ encodes the homotopy-theoretic information of $f:X\rightarrow Y$;
the standard triangulation on $K(\cal{A})$ encodes the homotopy information of the category of cochain complexes $C(\cal{A})$.

Any comments which might help to pin-down the motivation for the standard triangulation on $K(\cal{A})$ will be thankfully received.

REPLY [2 votes]: I am assuming this is not really news to anyone, but my take on triangules is that they provide a simple way to take "generalized quotients" of morphism. Here is what I mean by this: 
If there is a surjective morphism, adding the kernel provides a short exact sequence and similarly, for an injective morphism, adding the cokernel does the same.
Now with the cone construction we can do the same for any morphism. The beauty is that pretty much anything that one can do with regular short exact sequences (and the long exact sequences they induce) can be done using distinguished triangles.