Mapping cone of a composition Let f: $A \to B$ and $g: B \to C$ be morphisms of chain complexes of abelian groups. 
I have seen the statement now and then that the mapping cone $M_{g \circ f}$ of the composition is homotopy equivalent to $\text{Cone}(M_f \to M_g)$. 
I think I can prove this "by hand" (i.e. by writing down chain complexes), but I feel like there should be a more intrinsic/categorical proof.  Could someone sketch such an argument?
 A: This isn't quite the right statement, as one can easily check by letting $B=A\oplus C$ or by visualizing the situation for topological mapping cones. The correct statement is that, writing $M(f)$ for the mapping cone of $f$, we have a distinguished triangle $M(f)\to M(gf)\to M(g)$.
The intrinsic proof is necessarily higher categorical, because the cone of a morphism $f:A\to B$ represents the functor of morphisms out of $B$ equipped with a nullhomotopy of their restriction to $A$, a notion that can't be represented in a triangulated category. More generally, the cone of a morphism is not functorial on the triangulated category, which was almost immediately recognized as an important deficiency of the theory. 
The most down-to-earth way of fixing this issue is to consider the family of triangulated categories $D(\mathrm{Ab}^J)$ for various small categories $J$-this is the derived category of the abelian category of functors from $J$ to abelian groups. It is key to note that this is not by any means equivalent to the category of functors from $J$ into $D(\mathrm{Ab})$. The same problem of non-functoriality of the cone on $D(\mathrm{Ab})$, for instance, ensures that the arrow category $D(\mathrm{Ab})^\to$ is not even triangulated in any natural way. However, the cone is functorial on $D(\mathrm{Ab}^\to)$!
If $\square$ denotes the commutative square category and $*\leftarrow * \to *$ its subcategory with that shape, the restriction functor $D(\mathrm{Ab}^\square)\to D(\mathrm{Ab}^{*\leftarrow *\to *})$ has a left adjoint called the homotopy pushout functor. We can characterize $M(f)$ up to isomorphism in the derived category of arrows $D(\mathrm{Ab}^\to)$ as inhabiting a square 
$$\begin{matrix} A&\stackrel{f}{\to}&B\\\downarrow&&\downarrow\\0&\to&M(f)\end{matrix}$$ in the image of the homotopy pushout functor, in which the upper arrow is $f$ and the lower-left corner is zero. Then the composite $(A\to B)\mapsto (0\leftarrow A \to B)\mapsto \text {the square above}\mapsto C(f)$ gives the cone functor on $D(\mathrm{Ab}^\to)$, as promised.
Now, homotopy pushout squares satisfy the same pasting laws as ordinary pushout squares in categories; this follows from similar arguments, only rephrasing everything to rely solely on the adjunction relations between pushouts and restrictions. (See Groth's thesis for much more detail.) 
Therefore given a composable pair of arrows $A\stackrel{f}{\to}B\stackrel{g}{\to}C$, we may construct the following object of $D(\mathrm{Ab}^J)$, where $J$ is the category of three adjacent commutative squares as below, and all three squares are homotopy pushouts: 
$$\begin{matrix} A&\stackrel{f}{\to}&B&\stackrel{g}{\to}&C\\\downarrow&&\downarrow&&\downarrow\\0&\to&M(f)&\to&V\\&&\downarrow&&\downarrow\\&&0&\to&W\end{matrix}$$
By the pasting laws, also the wide rectangle with vertices $A,C,0,V$ is a homotopy pushout, which identifies $V=M(g\circ f)$. Similarly, the tall rectangle with vertices $B,0,C,W$ is a homotopy pushout, which identifies $W=M(g)$. Since the lower-right square is a pushout by construction, we have $M(g)=M(M(f)\to M(gf))$, as desired. 
