35 votes
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What is a triangle?

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 $...
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  • 37.1k
27 votes
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What is the correct definition of localisation of a category?

Actually, both of these definitions look weird to me. I would say there are two natural ways to define the localization $C[S^{-1}]$ by a universal property, as follows. For any category $D$, let ${\...
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  • 58.8k
20 votes
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Functorial kernel in derived category

Let $\mathcal{C}$ be a stable $\infty$-category. Then $\mathcal{C}$ has a homotopy category $h \mathcal{C}$, which is triangulated. The collection of morphisms $f: X \rightarrow Y$ of $\mathcal{C}$ ...
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  • 16.9k
19 votes
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idea and intuition behind triangulated category

It is risky to give a motivation for any concept in math and worse that of triangulated category that it is in a sense is a transitional concept form usual mathematics to mathematics up to homotopy. ...
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  • 8,183
18 votes

What (if anything) unifies stable homotopy theory and Grothendieck's six functors formalism?

The answer is motivic stable homotopy theory! This assigns to any reasonable scheme $S$ a triangulated category $SH(S)$ whose objects represent generalized cohomology theories for $S$-schemes, and ...
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18 votes
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Constructive homological algebra in HoTT

As regards HoTT, my own current opinion is that the best way to do "homological algebra" therein is by working directly with spectra. With only a working mathematician's knowledge of homological ...
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18 votes
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Recovering an abelian category from the Ext of its simple objects

Here's a counterexample that appears in nature. Fix a prime $p$ and a field $k$ of characteristic $p$, and let $G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if $p$ is odd, and $n\...
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16 votes
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Why is the derived tensor product only defined for bounded above derived categories?

I would like to add my 2 euro ¢. In a nutshell: your information is outdated. "Residues and Duality" by Hartshorne was the first available text on derived categories of coherent sheaves and ...
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  • 8,183
16 votes

What (if anything) unifies stable homotopy theory and Grothendieck's six functors formalism?

You can certainly form a category of sheaves of spectra on a space that is quite analogous to the derived category of sheaves of chain complexes. Roughly speaking, a presheaf of spectra is a called a (...
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  • 8,467
16 votes

So what exactly are perverse sheaves anyway?

If you're looking for a more geometric interpretation of perverse sheaves, you might be interested in MacPherson's 1990 lecture notes "Intersection Homology and Perverse Sheaves." As far as I know, I ...
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15 votes

Is there a (satisfying) proof that cellular cohomology is isomorphic to simplicial cohomology that doesn't use relative cohomlogy?

Here is how I like best to think about this, although I'll point to a few other proofs. Consider a general space $X$ (say compactly generated). There is a natural weak homotopy equivalence $\...
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  • 29k
15 votes
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When the restriction of a derived functor to a subcategory is the derived functor of the restriction

In the example where $\mathcal{D}$ is the category of abelian groups and $\mathcal{C}$ is the category of finite abelian groups, take $F(X)=X\otimes_\mathbb{Z}\mathbb{Q}/\mathbb{Z}$. Then the ...
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15 votes
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Example of an additive functor admitting no right derived functor

Let ${\cal C}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces equipped with a ${\bf Z}/2$ action, let ${\cal C'}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces and ...
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15 votes

Poincare duality on the level of complexes

One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology) $$...
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  • 15.6k
15 votes
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A concrete example of the deficiency of triangulated categories?

Since I have already given a similar answer recently, I don't want to be branded as the "anti-triangular" guy: the formalism of triangulated categories can be useful in certain settings. That said ...
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  • 15.6k
15 votes

Derived categories and classical theorems in homological algebra

Your question might be compacted to someting like: Do I need derived categories to study cohomology of sheaves? Of course, the answer depends on your particular interests. Let me anyway give you some ...
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  • 8,183
14 votes
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What is the applications of the dg-enhancements of derived categories of sheaves

It is hard to know where to begin! A general principle is that as long as you are only concerned with the derived category of a single variety, it is generally sufficient to consider it as a ...
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  • 5,559
14 votes

Recovering an abelian category from the Ext of its simple objects

This will only be possible when the abelian category $C$ is "Koszul" or formal in some sense. What will always be true is that the bounded derived category $D^{b}(C)$ (with its $dg$ or ...
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  • 1,180
13 votes
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An example of an object in $D^b_{\text{coh}}(\mathbb{P}^2)$ which is not formal

Yes; whenever you have two objects in an abelian category such that $Ext^2(M,N)$ is not equal to 0, we have a nonformal object given by coning with this morphism. More down-to-earthly, the element of ...
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  • 41.2k
13 votes
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Is the derived category of perfect complexes idempotent complete?

The derived category of perfect complexes is idempotent complete, because it is the sub category of compact objects in the derived category of quasi coherent sheaves (which is idempotent complete by ...
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  • 15.6k
13 votes

Why would one "attempt" to define points of a motive as $\operatorname{Ext}^1(\mathbb{Q}(0),M)$?

In the spirit of You Could Have Invented Spectral Sequences by T.Chow, I claim you could have invented $\operatorname{Ext}^{1}(\mathbb Q(0),M)$ as group of "rational points" of a motive. Here is how. ...
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  • 9,480
13 votes

What is a triangle?

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 ...
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13 votes
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Different definition of sheaf cohomology

Taking global sections is the same thing as computing the hom from $O_X$. In other words, there is an isomorphism of functors $\Gamma(X,-)\cong\hom(O_X,-)$, so both functors have the same derived ...
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13 votes
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Splitting of exact triangles in derived category

In any triangulated category, the necessary and sufficient condition for a distinguished triangle $A\to B\to C\to A[1]$ to split is that the morphism $C\to A[1]$ in this distinguished triangle ...
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13 votes
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Does formation of the derived $\infty$-category preserve pushouts?

A hands-on explanation: Relative tensor products like $B\otimes_AC$ are computed as the colimit of the simplicial object $B\otimes A^{\otimes \bullet} \otimes C$. The functor $\mathsf{Mod}_{(-)}: \...
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  • 12.7k
13 votes
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Embedding of a derived category into another derived category

Any fully faithful functor from $D^b(\mathcal{A})$ has adjoints (because $D^b(\mathcal{A})$ is a smooth and proper category), so its image is an admissible subcategory. A recent result from Dmitrii ...
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  • 31.5k
13 votes
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Computations in condensed mathematics, page 32-34

Correct, as both sides are the $S$-indexed direct sum of copies of $\mathbb{Z}$. For the LHS this holds by the universal property of $\mathbb{Z}[S]$, and for the RHS note that $C(S,\mathbb{Z}) = \...
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12 votes

What (if anything) unifies stable homotopy theory and Grothendieck's six functors formalism?

I feel like we gave up too quickly on parameterized spectra as a reasonable (if necessarily incomplete) answer to this question; we don't get the two formalisms in the question as special cases of an ...
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12 votes

What is the negative cyclic homology of a smooth projective variety?

There are conceptually simple definitions, but they require a more symmetric definition of Hochschild homology. The Hochschild homology of $X/k$ (with coefficients in $\mathcal O_X$) is the homology ...
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  • 7,644
12 votes
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Derived category of a quotient

What it is true is that $$ D^*(A/B) \simeq D^*(A)/D_B^*(A), $$ where the category $D_B^*(A)$ is the subcategory of $D^*(A)$ whose homologies lie in $B$. In general it may not agree with $D^*(B)$. ...
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  • 8,183

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