37
votes

Accepted

### 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 $...

28
votes

Accepted

### 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 ${\...

21
votes

Accepted

### 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 ...

21
votes

Accepted

### 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 $\mathcal{C}$ ...

20
votes

Accepted

### 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.
...

18
votes

Accepted

### 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\...

17
votes

Accepted

### Projective objects in the derived category of chain complexes

In a stable $\infty$-category, there are no nontrivial projectives. Of course, $0$ is always projective.
Now let $X$ be an arbitrary projective in some stable $C$, $X\simeq\Sigma \Omega X$ is a ...

17
votes

Accepted

### Is the functor from the unbounded derived category of coherent sheaves into the derived category of quasi-coherent sheaves fully faithful?

No, not always.
In
Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID ...

16
votes

Accepted

### 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 ...

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 ...

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 $\...

15
votes

Accepted

### 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 ...

15
votes

Accepted

### 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 ...

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)
$$...

15
votes

Accepted

### 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 ...

15
votes

Accepted

### 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 ...

14
votes

Accepted

### 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 ...

14
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 ...

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 ...

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.
...

13
votes

Accepted

### 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 ...

13
votes

Accepted

### 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 ...

13
votes

Accepted

### Derived Category of the derived critical locus, is it the category of Matrix Factorizations?

These are indeed related. The first thing to know is that they both ``live'' (i.e., sheafify) over the critical locus (this is not saying much if you assume $W$ has isolated critical points, but ...

13
votes

Accepted

### 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}_{(-)}: \...

13
votes

Accepted

### 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 ...

13
votes

Accepted

### 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}) = \...

12
votes

Accepted

### Is there a compact generated triangulated category which does not have a compact generator?

Examples can be found amongst the derived categories of algebraic stacks, see Hall--Rydh: Algebraic groups and compact generation of their derived categories of representations, in particular theorem ...

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 ...

12
votes

Accepted

### 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 $D^*(B)$.
...

12
votes

### Determinantal identities for perfect complexes

The formula also holds for perfect complexes. This can be deduced from the case of vector bundles, although it requires a lot of structure in that case. Namely, we need to use the fact that the ...

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