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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 …
Yonatan Harpaz's user avatar
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, …
Yonatan Harpaz's user avatar
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 …
Yonatan Harpaz's user avatar
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 …
Yonatan Harpaz's user avatar
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 …
Yonatan Harpaz's user avatar