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

  1. the standard triangle $ X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $ encodes the homotopy-theoretic information of $f:X\rightarrow Y$;

  2. 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.


4 Answers 4


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.


It certainly yields intuition to look at chain complexes of singular simplices, coming from topological spaces: Look at the nice table on page 16 of these notes of Behrang Noohi and do the exercise he suggests there (actually the whole second chapter is about your question).

This tells you exactly what is going on for chain complexes concentrated in positive degrees, as these always correspond to spaces (via the Dold-Kan theorem). For general chain complexes it would then pay off, in terms of intuition, to study the beginnings of stable homotopy theory - pages 15-23 of Bjørn Dundas' notes give a brief idea, the first chapter of Adams' "Infinite Loop spaces" is another source for this which focuses on intuition.

Finally there are abstract frameworks describing situations (i.e. categories) where a notion of homotopy exists and they capture the essence of both the algebraic and the topological examples. The currently most popular such framework are "model categories" (chain complexes are a so-called "stable" model category). There the notions of homotopy kernel and homotopy cokernel have precise meaning and you can see that mapping cones play that role in the category of chain complexes. The friendliest starting point for this is Dwyer/Spalinski's "Homotopy theories and model categories" - chain complexes are a main example there.


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.


One possible motivation for considering the triangles in 1. is that they induce long exact sequences
$\cdots \rightarrow Hom_K(Z,C_f[-1]) \rightarrow Hom_K(Z,X) \rightarrow Hom_K(Z,Y)\rightarrow Hom_K(Z,C_f) \rightarrow \cdots$
$\cdots \rightarrow Hom_K(C_f,Z) \rightarrow Hom_K(Y,Z)\rightarrow Hom_K(X,Z) \rightarrow Hom_K(C_f[-1],Z) \rightarrow \cdots$
for any $Z$.

I think that the triangulated structure of $K(\mathcal{A})$ reflects (in some sense "up to homotopy") the abelian structure of $C(\mathcal{A})$. Indeed $K(\mathcal{A})$ is the stable category (see the book of D. Happel "Triangulated categories in the representation theory of finite dimensional algebras") associated with the abelian category $C(\mathcal{A})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.