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Results tagged with derived-functors
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user 51164
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 l …
- 9,491
10
votes
Accepted
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 ki …
- 9,491
6
votes
Accepted
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 w …
- 9,491
2
votes
Commutativity between functors on sheaves of abelian groups
1) No. For example, if $A$ is non-empty and $f: X \to \ast$ is the map to the point then $j^*R^qf_*\mathbb{C} \cong H^q(X,\mathbb{C})$ while $R^q(f|_A)_*i^*\mathbb{C} \cong R^q(f|_A)_*\mathbb{C} \cong …
- 9,491
1
vote
Deriving the functor $ \int_{\Gamma} F(-,-)$
The answer is no in this generality, but I do not know what happens specifically for ${\bf dgCat}$ and ${\bf dgFun}$. For convenience, let me construct a counter-example to the dual of your question, …
- 9,491