So I've been reading about derived categories recently (mostly via Hartshorne's Residues and Duality and some online notes), and while talking with some other people, I've realized that I'm finding it difficult to describe what a "triangle" is (as well as some other confusions, to be described below).

Let $\mathcal{A}$ be an abelian category, and let $K(\mathcal{A})$ and $D(\mathcal{A})$ be its homotopy category and derived categories respectively.

By definition, in either $K(\mathcal{A})$ or $D(\mathcal{A})$,

(1) A triangle is a diagram $X\rightarrow Y\rightarrow Z\rightarrow X[1]$ which is isomorphic to a diagram of the form $$X\stackrel{f}{\rightarrow}Y\rightarrow Cone(f)\rightarrow X[1]$$

This is the definition, though I don't really understand the motivation.

Somewhat more helpful for me, is the definition:

(2) A cohomological functor from a triangulated category $\mathcal{C}$ to an abelian category $\mathcal{A}$ is an additive functor which takes triangles to long exact sequences.

Since taking cohomology of a complex (in either $K(\mathcal{A})$ or $D(\mathcal{A})$) is a cohomological functor, this definition tells me that I should think of a triangle as being like "a short exact sequence" (in the sense that classically, taking cohomology of a short exact sequence results in a long exact sequence). This idea is also supported by the fact:

(3) If $0\rightarrow X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow 0$ is an exact sequence of chain complexes, then there is a natural map $Z^\bullet\rightarrow X[1]^\bullet$ in the derived category $D(\mathcal{A})$ making $X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow X[1]^\bullet$ into a triangle (in $D(\mathcal{A})$).

This leads to my first precise question: Can there exist triangles in $D(\mathcal{A})$ which don't come from exact sequences? If so, is there a characterization of them? Is (3) false in the homotopy category $K(\mathcal{A})$? (certainly the same proof doesn't work).

I sort of expect that the answers to the first and last questions above to both be "Yes", which makes the comparison between triangles and "exact sequences" a bit weird. Of course, $K(\mathcal{A})$ and $D(\mathcal{A})$ are almost never abelian categories, and so it's "weird" to talk about exact sequences there.

I suppose at a fundamental level, I find the "homotopy category" somewhat mysterious. I don't completely understand it's role in the construction of the derived category, since after all homotopy equivalences are quasi-isomorphisms. I also find it difficult to internalize this notion of "homotopic morphisms of chain complexes". To me, it's just a "technical trick" which allows one to do all this magic with mapping cones which allows $K(\mathcal{A})$ to be a triangulated category, whereas the normal abelian category of chain complexes is not. I can prove things with it, but whenever I do, I sort of feel unsettled - as if I'm playing with something 'magical' that could, at any moment, turn on me unexpectedly.

In addition to my specific questions above, I suppose I was hoping that someone would be able to articulate in a nice way how one should think of triangles, why this notion of a triangulated category is so successful, and hopefully alleviate some of my unsettlement regarding homotopy.

  • 1
    $\begingroup$ Isn't the mapping cone always a long exact sequence explicitly, so has to come by this construction? Is your problem things that arise by long exact sequences that aren't long exact sequences? $\endgroup$
    – Will Sawin
    Dec 18, 2016 at 11:10
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    $\begingroup$ I believe the point of this mod out by homotopy thing is that homs between two complexes in the derived category really want to be a complex themselves (i.e. we should all be working in the dg category). The naive homs in the category of complexes correspond to taking the kernel of $d$ from degree $0$ to degree $1$ in this complex, and homotopy is modding out by the image of $d$ from degree $-1$ to degree $0$, calculating the $0$th homology. $\endgroup$
    – Will Sawin
    Dec 18, 2016 at 11:14
  • $\begingroup$ @WillSawin Ah, of course mapping cones already give rise to long exact sequences! Nice! I'll have to think about your second comment more... $\endgroup$
    – Will Chen
    Dec 19, 2016 at 5:14
  • $\begingroup$ Are these homological triangles somehow related to the geometric triangles of Euclid? Or is the name "triangle" simply a reference to the shape of the diagram? (Example: I think category-theoretic cones are generalizations/abstractions of geometric/topological cones.) $\endgroup$ May 22, 2017 at 6:28

3 Answers 3


To answer your first precise questions:

  • Yes, every distinguished triangle in $D(A)$ comes from a short exact sequence. For every distinguished triangle $X \to Y \to \mathrm{Cone}(f) \stackrel{+1}\to $ there is a short exact sequence $$ 0 \to Y \to \mathrm{Cone}(f) \to X[-1] \to 0$$ of complexes in $A$, and our distinguished triangle arises from this one by rotation.

  • By the same argument, every distinguished triangle in $K(A)$ comes from a short exact sequence (at least up to a rotation). However, not every short exact sequence gives rise to a distinguished triangle in $K(A)$. If $$ 0 \to X \stackrel f \to Y \to Z \to 0$$ is a short exact sequence of complexes in $A$, then there is a natural map $\mathrm{Cone}(f) \to Z$ which in general is only a quasi-isomorphism, not a homotopy equivalence. If for example the short exact sequence splits degree-wise, then it is always a homotopy equivalence, and we get a triangle in $K(A)$.

Alright. About your question "Why $K(A)$?". You are right that homotopy equivalences are quasi-isomorphisms. So in principle one could construct $D(A)$ in "one step" by taking the category of complexes and inverting quasi-isomorphisms. But there are some technical reasons for preferring the construction via the category $K(A)$. First off, the category $K(A)$ is already triangulated, and this is easy to prove. This means that to construct $D(A)$ we are in the situation of Verdier localization: we have a triangulated category, and we localize it at the class of morphisms whose cone is in a specific thick triangulated subcategory. In particular $D(A)$ becomes triangulated. In general, localizations of categories can be complicated, and in the "one-step" construction it is not even obvious that $D(A)$ is a category, i.e. that the morphisms from one object to another form a set.

To motivate why we should care about triangles at all, note that it doesn't make sense to talk about kernels or images of morphisms in $D(A)$, so that we can't talk about exactness. Given that we want to be able to do homological algebra we need some sort of substitute for this. Triangles, encoding short exact sequences, turn out to be enough to develop much of the theory, and a posteriori one could want to axiomatize the theory only in terms of triangles. One reason we could expect the notion of a "distinguished triangle" to really be intrinsic to $D(A)$ (even though the class of distinguished triangles need to be specified in the axioms for a triangulated category) is that any triangle $$ X \to Y \to Z \stackrel{+1}\to$$ gives rise to a long sequence of abelian groups after applying $[W,-]$ for any object $W$; this long sequence will be a long exact sequence when the triangle is distinguished, for any $W$, and this is really a very special property! A remark is that nowadays people will tell you that "stable $\infty$-categories" are for all purposes better than triangulated categories, and in a stable $\infty$-category, the class of distinguished triangles does not need to be specified in advance, so to speak: equivalent stable $\infty$-categories will have the same distinguished triangles.

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    $\begingroup$ Great answer. I just want to add that I personally found that switching terminology and calling them "exact triangles" instead of "distinguished triangles" made it much easier for me to understand derived categories when I was first learning them. Hopefully this is helpful to others as well. $\endgroup$ Jan 19, 2017 at 20:09

You really seem to be looking for intuition for the triangulated structures on derived categories of Abelian categories, so here goes:
(Co-)chain complexes are like (Abelianised) pointed homotopy types; homotopy types are typically first encountered in the guise of topological spaces, so it is in the category $\mathbf{Top}_*$ of pointed topological spaces that our story begins. Pointed homotopy types do not naturally arrange into a category, but an $(\infty,1)$-category (I will just write $\infty$-category); this is a generalisation of the notion of a category. The data given by $\mathbf{Top}_*$ together with the subcategory of weak homotopy equivalences completely determines this $\infty$-category. (Conceptually, you just universally invert the weak homotopy equivalences in the $\infty$-category of $\infty$-categories; concretely you could consider the simplicial localisation.)
Just like in ordinary categories you can define limits and colimits in $\infty$-categories. The limits and colimits in the $\infty$-category of homotopy types (indexed by ordinary small categories) can be computed directly in $\mathbf{Top}_*$; one typically calls such a construction a homotopy (co-)limit. Given a continuous map $f: X \to Y$ between pointed spaces, the homotopy pushout of $f$ along the unique map $X \to 0$ is called the cofibre of $f$. An easy construction of this homotopy colimit is given by the mapping cone $C_f$ of $f$. You can now take the cofibre of $Y \to C_f$ to obtain the diagram $$\require{AMScd} \begin{CD} X @>>> Y @>>> 0 \\ @VVV @VVV @VVV \\ 0 @>>> C_f @>>> \Sigma X. \end{CD}$$ As indicated in the diagram we see that "taking the cofibre twice" of any continuous maps with domain $X$ yields the suspension of $X$; this is because homotopy pushouts compose, and the cofibre of $X \to 0$ is $\Sigma X$. (If this is new to you, I recommend working through some examples, e.g. in here.) We can now iterate the cofibre construction of $f$ to obtain the cofibre sequence of $f$.
If we apply a given cohomology theory $\{H^n\}$ to this cofibre sequence, we obtain the sequence $H^n(X) \leftarrow H^n(Y) \leftarrow H^n(C_f) \leftarrow H^{n-1}(X) \leftarrow \cdots$. If $X \to Y$ is an inclusion, then $H^i(C_f) \cong H^i(X,Y)$. We see that cohomology theories take cofibre sequences to long exact sequences; this together with characterises cohomology theories. We also see that for $n > m$, we can reconstruct $H^m$ from $H^n$, so it feels as if we only had had one functor, which takes cofibre sequences to long exact sequences of Abelian groups. The main problem is that we can't invert $\Sigma: \mathbf{Top}_* \to \mathbf{Top}_*$. If you universally (in the $\infty$-categorical setting) construct a category from $\mathbf{Top}_*$ in which $\Sigma$ is invertible, you obtain the category of spectra, and any cohomology theory on spectra is indeed determined by $H^0$. Spectra are typically thought of as some sort of "Abelianised" homotopy types.
We shall however not consider spectra, but something similar yet much simpler: chain complexes! First, all of the above discussion could have taken place in the category $\mathbf{SSet}_*$ of pointed simplicial sets. This category also has a subcategory of weak equivalences, which for nice objects (Kan complexes) look exactly like weak homotopy equivalences of topological spaces. Again, $\mathbf{SSet}_*$ together with its weak equivalences presents the $\infty$-category of pointed homotopy types (this is typically expressed by the fact that the total singular complex functor $\mathrm{Tot:} \mathbf{Top}_* \to \mathbf{SSet}_*$ is part of a Quillen equivalence). We said that spectra could be thought of as "Abelianised" homotopy types; in $\mathbf{SSet}_*$ there is an easy way of approximating what is meant by this as follows: There is a free forgetful adjunction $\underline{\;\;} \otimes \mathbb{Z}: \mathbf{SSet}_* \rightleftarrows \mathbf{SAb}:U$ between pointed simplicial sets and simplicial Abelian groups. Let $\mathbf{Ch}_{\geq 0}(\mathbf{Ab})$ denote the category chain complexes of Abelian groups concentrated in non-negative degree; then (a variant of) the functor taking alternating sums of face maps gives an equivalence of categories $\mathbf{SAb} \xrightarrow{\simeq} \mathbf{Ch}_{\geq 0}(\mathbf{Ab})$. This is the Dold-Kan equivalence. Given a simplicial Abelian group, then the homology groups of the associated chain complexes are canonically isomorphic to the homotopy groups of $X$. (Note that we can compose the three functors discussed above to obtain a functor $\mathbf{Top}_* \to \mathbf{Ch}_{\geq 0}(\mathbf{Ab})$; composing this functor with $n$-th homology just gives you the $n$-th singular homology.) We now see that quasi-isomorphisms correspond exactly to weak homotopy equivalences of spaces. Furthermore $\mathbf{Top}_* \to \mathbf{Ch}_{\geq 0}(\mathbf{Ab})$ takes homotopies between continuos maps to homotopies of chain complexes.
Again chain complexes naturally arrange into an $\infty$-category; again, this encoded by the fact that we have a subcategory, this time the quasi-isomorphisms, that we would like to invert. Let us embed $\mathbf{Ch}_{\geq 0}(\mathbf{Ab})$ into the category $\mathbf{Ch}(\mathbf{Ab})$ of bounded or unbounded chain complexes. Here we can again consider cofibre sequences: Let $f: X \to Y$ be a map of chain complexes, then the cofibre of $f$ can again be constructed using the mapping cone, and taking the cofibre twice in this case yields a chain complex quasi-isomorphic to $X[1]$ (Again, $\mathbf{Top}_* \to \mathbf{Ch}_{\geq 0}(\mathbf{Ab})$ takes the suspension of a spaces to a chain complex quasi-isomorphic to the shift of the associated chain complex.). A (co-)homological functor can then be defined as a co(-ntra-)variant functor from $\mathbf{Ch}(\mathbf{Ab})$ to Abelian groups taking cofibre sequences to long exact sequences of Abelian groups. Let $A$ be an Abelian group, to construct examples of (co-)homological functors corresponding to ordinary (co-)homology with coefficients in $A$, consider the functors $H_0 \circ \underline{\; \;}\otimes \mathbb{Z}$ and $H_0 \circ \mathrm{Hom}(\underline{\;\;},A)$ respectively.
So what about short exact sequences of chain complexes? It turns out that if $f: X \to Y$ is a monomorphism, then the ordinary quotient $Y/X$ already is the cofibre of $f$! The long exact sequences $\cdots H^1(Y/X) \to H^0(X) \to H^0(Y) \to H^0(Y/X) \to H^{-1}(X) \to \cdots$ using the snake lemma is (canonically) isomorphic to the sequence $\cdots H^1(C_f) \to H^0(X) \to H^0(Y) \to H^0(C_f) \to H^{-1}(X) \to \cdots$ obtained from the cofibre sequence.
Now, you might have noticed something fishy going on; I didn't tell you where the morphism $H^1(C_f) \to H^0(X)$ came from! It turns out that $\Sigma$ can be defined on any pointed $\infty$-category (with finite limits and colimits), and that requiring this functor to haven an inverse is the same as requiring that any commutative square is a pushout square iff it is a pullback square. Such $\infty$-categories are called stable $\infty$-categories. In such an $\infty$-category any morphism $X \to Y$ can be extended in both directions, by cofibre sequences to the right and by fibre sequences to the left!
To finally get to the triangulated category of an Abelian category $\mathcal{A}$, we note that from any $\infty$-category $\mathcal{C}$ we can universally construct an ordinary category; the so-called homotopy category $\mathrm{Ho}(\mathcal{C})$ of $\mathcal{C}$. The homotopy category of the stable $\infty$-category presented by the category of (co-)chain complexes in $\mathcal{A}$ (with quasi-isomorphisms) is $D(\mathcal{A})$. Any morphism $X \to Y$ in $\mathbf{Ch}(\mathcal{A})$ is the input needed to construct a cofibre sequence (or now equivalently fibre sequence), but we are no longer able construct this cofibre sequence, once we have descended to $D(\mathcal{A})$; the data of the triangulated structure on $D(\mathcal{A})$ is exactly what is needed to retain this information, and how different cofibre sequences relate to each other. In fact, it turns out that the homotopy category of any stable $\infty$-category canonically admits the structure of a triangulated category. It now appears completely natural that we should call those functors (co-)homological which take distinguished triangles to long exact sequences.
As for your precise question, any morphism in $\mathbf{Ch}(\mathcal{A})$ can be replaced with its mapping cylinder, which is a monomorphism, so any to adjacent morphisms in a long exact sequence may be presented by a short exact sequence of (co-)chain complexes, so distinguished triangles in $D(\mathcal{A})$ all come from short exact sequences. This is almost certainly not true in $K(\mathcal{A})$, but I don't have any particular counterexample in mind.


This is not an answer to your question, but I think that you'll find this piece of information to be useful:

A triangle in a stable infinity category is given by finitely many pieces of data:

• Three objects: $A$, $B$, $C$.

• Three morphisms:
$f:A\to B$,
$g:B\to C$,
$h:C\to A[1]$

• Two 2-morphisms:
$\alpha:0\to gf$
$\beta:kg \to 0$

The condition that these objects, morphisms, and 2-morphisms need to satisfy is that the loop $h\alpha\cup\beta f\in \Omega Hom(A,A[1])=Hom(A,A)$ (known as a Toda bracket)
should be homotopic to the element $id\in Hom(A,A)$.

A very nice consequence of the fact that triangles can be expressed in the above simple way is that any functor between stable infinity categories sends triangles to triangles.

In that sense, triangles in stable infinity categories are rather different from short exact sequences in abelian categories.


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