26 votes
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Existence and uniqueness of Haar measure on compacta; a cohomological approach

Fix a compact group $G$ and consider its category of Banach representations: the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such ...
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  • 10k
18 votes
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Abelian category with enough injectives but not functorially

Since the dual of an abelian category is also an abelian category, the question is equivalent to the same question for projective resolutions. I will show that the category $\mathbf{Ab}^{\...
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16 votes
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Unifying "cohomology groups classify extensions" theorems

$\newcommand{\cA}{\mathcal{A}}\newcommand{\Ext}{\mathrm{Ext}}\newcommand{\Hom}{\mathrm{Hom}}$Let $\cA$ be an abelian category; then, $\Ext_\cA^i(A,B)$ is literally $\Hom_{D(\cA)}(A, B[i])$, where $B[i]...
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  • 5,133
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
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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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14 votes
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Derived functors - homotopical vs homological approach

EDIT Corrected a couple of inaccuracies and mistakes, added some references. For the sake of clarity, let me work with non-negatively graded cochain complexes, and analyze the case of a left exact ...
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12 votes

Grothendieck spectral sequence when one of the functors is contravariant

I think this case is actually not so obvious. The issue is that to derive $R\mathscr Hom$ in the first variable you would need to use a locally free resolution while $Rf_*$ being a covariant right ...
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9 votes
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Fourier-Mukai functors being identity on objects

If $X$ is smooth and projective, then any such FM functor is in fact naturally isomorphic to the identity functor. This follows immediately from Corollary 5.23 of Huybrechts' book on Fourier-Mukai ...
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9 votes
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Different definitions of derived functors

The total right derived functor ${\bf R}F(-)$ contains a bit more information than just its individual cohomologies ${\bf R}^iF(-) = H^i({\bf R}F(-))$. This information can indeed be described as a ...
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8 votes
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The Mittag-Leffler condition as necessary and sufficient

This is due to Emmanouil as far as I know. See this.
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  • 96
8 votes
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Grothendieck-Verdier duality without the noetherian condition

Does one have the duality in this setting? Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward ...
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  • 2,736
8 votes
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Can we define derived functors in model categories without functorial factorisations?

This depends on whether one insists on derived functors landing in the original model category (as is the case with modern approaches of Hinich and Dwyer–Hirschhorn–Kan–Smith), or in its homotopy ...
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7 votes
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Question on condition for a sheaf to be locally free in Orlov 2004

The question is local, so it is enough to show that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ and $M$ is a finitely generated module such that $Ext^i(M,A/\mathfrak{m}) = 0$ ...
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  • 31.9k
7 votes
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Two basic questions on derived categories

First of all, the functor $Rf_\ast$ is a red herring. Take an injective resolution $I^{\bullet,\bullet}$ of $F^\bullet$ and apply $f_\ast$; then you just have two questions regarding double complexes. ...
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  • 37.5k
7 votes

What is integration along the fibers in D-module theory?

I think, for a locally constant $D$-module it basically cashes out to Defining a formal solution to the differential equation. Defining a formal integral along the fibers. Writing down all the ...
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  • 117k
7 votes
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Derived Nakayama for complete modules

Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are: For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\...
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6 votes
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Derived pullback of the coarse moduli morphism

If both $\mathcal{X}$ and $X$ are locally Noetherian and regular, then $f$ is flat. Then $Lf^*$ is the usual pullback $f^*$. If $\mathcal{X}$ is tame, then the natural transformation $$\theta:\text{...
6 votes

Motivation/intuition behind the definition of delta-functors and related concepts

To be honest, the question is a bit borderline for MO, but I'll make a few comments anyway. I would argue that there is nothing inherently geometric about the definition of (universal) delta functors, ...
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  • 31.9k
6 votes
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Questions about $\text{Perf}(A)$ of dg algebra $A$

As explained a little bit further in Elagin and Lunts' paper, the category $Perf(A)$ consists of the compact objects of $D(A)$, this is exactly what happens in the usual situation in algebraic ...
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  • 599
6 votes
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How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$?

The cone of $i^*i_*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}_X(-X)[2]$. EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle $$ i^*...
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  • 31.9k
6 votes
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Is the composite of absolute derived functors a derived functor?

Here is a somewhat degenerate example that illustrates what can go wrong. Let $\textbf{Ab}$ be the category of abelian groups, considered as a homotopical category where the weak equivalences are the ...
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  • 13.3k
5 votes
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The naive approach to deriving profunctors - What's wrong with it?

The problem with this definition is that the formula $\mathrm{colim}_{X \to Z \in \mathcal{W}} F(Z)$ does not, in general, depend functorially on $X$. For it to depend functorially on $X$ you need a ...
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5 votes
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$Lf^*$ is fully faithful

Smoothness is not necessary. What is important is that $f$ has finite $Tor$-dimension (otherwise $Lf^*$ does not preserve boundedness); a sufficient (but not necessary) condition for this is ...
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  • 31.9k
5 votes
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multiplicative structure of Ext

So, maybe I should format this as an answer. Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(...
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5 votes
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Subspace inclusion with non-vanishing higher direct images

You expect the higher direct images of $f: X \to Y$ to be nonzero if for arbitrarily small neighborhoods $y_0 \in U$ the space $f^{-1}(U)$ has non-vanishing higher cohomology. I.e. $R^i f_* \mathbb Z$...
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5 votes
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Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication

Remark. Exact sequence $0 \to L \to E \to M \to 0$ corresponds to $Ext^1(M,L)$, not to $Ext^1(L,M)$. Q1. $a \in k^\times$ acts on $Ext^1(L,M)$ via pullback along $a:L \to L$ or via pushout along $a: ...
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  • 31.9k
5 votes

Conflicting definitions of RHom

The bifunctor $$R\operatorname{Hom}^\bullet \colon D(\mathcal{A}) ^{op} \times D(\mathcal{A}) \to D(\operatorname{Ab}) $$ is understood —whenever $\mathcal{A}$ has enough injective objects (e.g. $\...
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  • 8,261
5 votes
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Derived functor of functor tensor product

The answer is yes if you assume enough things. In particular, the notion of a left flat object of $\mathcal A$ comes up : Definition: An object $L\in\mathcal A$ is left flat if $-\otimes L$ is exact. ...
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  • 8,860
4 votes

Grothendieck spectral sequence when one of the functors is contravariant

If $I^*$ and $J^*$ are global sections of appropriate injective resolutions of $f_*{\cal O}_X$ and ${\cal O}_S$, then you have a double complex that looks in part like this: $$\matrix{ Hom(I^q,J^p)&...
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4 votes
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Balanced dualizing complexes according to A. Yekutieli

I happened to be looking at this very issue a few days ago! Here is my explanation to myself... Indeed, in the notation above, there is a homomorphism of complexes of graded modules (not derived) $...
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