Exact triple yields a distinguished triangle in derived category

In Methods of Homological Algebra before Proposition III.3.5 there is a short comment: "The next proposition shows that any exact triple [of complexes] can be competed to a distinguished triangle".

How should one understand this comment?

In other words, given exact sequence of complexes in abelian category $\mathcal A$:

$$0\rightarrow K\stackrel{f}\rightarrow L\rightarrow M\rightarrow 0$$

with no conditions on maps and complexes, is there always a way to construct a map $M\rightarrow K[1]$ and isomorphism (in the category $\operatorname {Com}(\mathcal A)$) of the obtained triangle with the distinguished triangle

$$\rightarrow K\rightarrow \operatorname{Cyl} (f)\rightarrow \operatorname{Cone}(f)\rightarrow K[1]\rightarrow$$

Proposition III.3.5 is a construction of quasi-isomorphism between the exact triple and $0\rightarrow K\rightarrow \operatorname{Cyl} (f)\rightarrow \operatorname{Cone}(f)\rightarrow 0$, but how is that related to the comment?

• You are right. I think by "completing" the author meant adding the morphism $M \to K[1]$. Commented May 15, 2014 at 3:21

I think the intended meaning is that any exact triple in $\mathrm{Com}(A)$ can be completed to a distinguished triangle in $D(A)$. So the map $M \to K[1]$ is the composition $M \overset{\gamma^{-1}}{\longrightarrow} C(f) \overset{\delta}{\longrightarrow} K[1]$, where $\gamma$ is the quasi-isomorphism constructed in the proof. Note that $\gamma^{-1}$ is only a map in $D(A)$, not in $\mathrm{Com}(A)$.

It is easy to give an example where the map $M \to K[1]$ doesn't exist in $\mathrm{Com}(A)$. (I changed the example to make correctness easier to verify.) For example, let $A$ be the category of $k[x]$-modules, and let $K$, $L$ and $M$ be the complexes which are concentrated in degree $0$, with $0 \to K_0 \to L_0 \to M_0 \to 0$ given by $$0 \to k[x] \overset{x}{\longrightarrow} k[x] \to k \to 0$$.

Clearly, the only map $M \to K[1]$ in $\mathrm{Com}(A)$ is the zero map. I claim that there is no commuting diagram $$\begin{matrix} \mathrm{Cone}(f) & \longrightarrow & K[1] \\ \downarrow & & \downarrow \\ M & \overset{0}{\longrightarrow} & K[1] \\ \end{matrix}$$ in $D(A)$, with the vertical maps isomorphisms. Thus, if we choose $M \to K[1]$ to be zero, we will not be able to make the resulting triangle isomorphic with the distinguished triangle.

It is enough to show that $\mathrm{Cone}(f) \to K[1]$ is not zero in the derived category (although it induces the $0$ map on cohomology.) Since the modules in $\mathrm{Cone}(f)$ are projective, it is enough to check that this map is not null-homotopic. (Exercise III.5.1 -- in my opinion, it is criminal of this book to leave this crucial fact as an exercise, but there it is.) That is straightforward, and I leave it as an exercise.

Here is the relevant page in google books.

• Is $A$ the category of $R$-modules in your example? Isn't it the case that we can take $M \stackrel 0\rightarrow K[1]$? It seems, that corresponding long exact sequence works fine then. Commented May 15, 2014 at 18:21
• You shouldn't be able to fit that into a commuting diagram of quasi-isomorphisms with the distinguished triangle coming from the cone and cylinder. I am pretty sure I worked out at some point that $M \overset{0}{\to} K[1]$ is is quasi-isomorphic to the triangle coming from the cone/cylinder construction if and only if the underlying sequence of modules is split in $A$, and this one isn't. Commented May 15, 2014 at 18:41
• Edited to sketch more of the argument. Commented May 16, 2014 at 14:16