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 $f:X\rightarrow Y$;
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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.
Best Answer
The motivation comes from topology. Suppose you have a map of topological spaces $f:X \to Y$. You want to form the homotopical version of $Y/X$, so instead of identifying X to a point immediately, you do so gradually, so you form $Y \cup_f CX$, where X is the cone on X, and you glue the base of the cone to Y by the map f.
So from the point of view of the cells, the cells of the mapping cone on f in dimension n are
the cells of Y in dimension n
the cells of X in dimension n-1 (cross the 1-cell in the cone direction).
To do this with cochain complexes, you ought to form
Y in degree n
direct sum
X in degree n+1 (because you're using cochain, not chain)
The differential also comes from the topological motivation.