Start with A an abelian category and form the derived category D(A).
Take a triangle (not necessarily distinguished) and take it's cohomology. We obtain a long sequence (not necessarily exact). If the triangle is distinguished it is exact. How about the converse: if the long sequence in cohomology is exact does it follow that the triangle is distinguished? (My guess is no, but I can't find a counter-example).
[Math] distinguished triangles and cohomology
derived-categorieshomological-algebratriangulated-categories
Related Solutions
First, consider the category $\mathcal{S}$ of spectra. Let $S/2$ denote the cofibre of twice the identity map of the sphere spectrum. It is then known that the identity map of $S/2$ has order $4$. If we had a good system of chain complexes $Hom_{\mathcal{S}}(x,y)$ then $Hom_{\mathcal{S}}(S/2,S/2)$ would be the cofibre of twice the identity on $Hom_{\mathcal{S}}(S/2,S)$, but for chain complexes the cofibre of twice the identity always has exponent 2, which gives a contradiction. Thus, for triangulated categories of topological origin, the best you can hope for is to define mapping spectra $F(x,y)$, not chain complexes $Hom(x,y)$.
Next, this paper:
- Fernando Muro, Stefan Schwede, Neil Strickland, Triangulated categories without models, Inventiones Mathematicae, Vol. 170 (2007), No. 2, pp. 231-241, https://doi.org/10.1007/s00222-007-0061-2, https://arxiv.org/abs/0704.1378
exhibits a triangulated category $\mathcal{C}$ such there are no nontrivial exact functors from $\mathcal{C}$ to any of the standard examples of triangulated categories, or in the opposite direction. This means in particular that there is no way to define a 'representable' functor $F_{\mathcal{C}}(x,-)$.
However, if $\mathcal{D}$ is a triangulated category arising from a stable model category then one can define mapping spectra $F_{\mathcal{D}}(x,y)$ for objects $x,y\in\mathcal{D}$. I think that is covered by this paper:
- Stefan Schwede, Brooke Shipley, A uniqueness theorem for stable homotopy theory, Mathematische Zeitschrift 239 (2002), 803-828, https://doi.org/10.1007/s002090100347, https://arxiv.org/abs/math/0012021.
As in David Roberts's comment, I think one can also do the same for triangulated categories that arise from stable $(\infty,1)$-categories.
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.
Best Answer
An important property of the derived category is that distinguished triangles don't just produce long exact sequences in cohomology. If A -> B -> C -> A[1] is an exact triangle and E is another object in the derived category, then you get a long exact sequence
where these Hom-sets are sets of maps in the derived category.
A particular counterexample is as follows. We can view any abelian group as a chain complex concentrated in degree zero. There is a distinguished triangle as follows:
However, we can take the last map ℤ/2 -> ℤ[1] in the sequence (which is not zero in the derived category) and replace it with the zero map. This still gives us a long exact sequence on (co)homology groups. However, if we let E = ℤ/2, then applying maps in the derived category from our new non-distinguished triangle gives us the sequence
where the last map is induced by the zero map from ℤ/2[-1] to ℤ, and so it must be zero. This sequence can't possibly be exact, and so the new triangle is not distinguished.
EDIT: A more subtle question this suggests is: "Suppose I have a triangle and, for any E, applying Map(-,E) or Map(E,-) gives a long exact sequence. Is this a distinguished triangle?"
The answer to this is actually still no. Still considering chain complexes of abelian groups, take the distinguished triangle
where I'll call the last map β (for Bockstein). You can take this distinguished triangle and replace β with its negative -β. The new triangle still induces long exact sequences on maps in or maps out (because the maps in the triangle have the same kernel and image). However, as an exercise show that this can't be a disntinguished triangle because it's not isomorphic to the original distinguished triangle.