# Cohomology theory “from” Grothendieck's six operations?

How, precisely, (as suggested by grothendieck) cohomology theory naturally follows from the Grothendieck's six operations associated to the category derived from the topos?

I would like some references.

• The most general reference I know of is this: arxiv.org/abs/0912.2110, but I hope someone will show up and give a more beginner-friendly one. – Denis Nardin May 6 '16 at 2:44
• I don't think I can parse your question, but cohomology is (derived) pushing forward to a point- which is one of the 6 operations. – Dylan Wilson May 6 '16 at 13:27
• @DylanWilson I am interpreting the question by asking to reduce the usual properties of a cohomology theory (e.g. duality, projection formula, cycle class map...) to the formalism of six operations rather than just defining the cohomology theory. – Denis Nardin May 6 '16 at 15:01

Assuming you have the usual 6 operations we have the following recipes for co/homologies of $f \colon X \to Y$ with coefficients $\mathcal{F}$ over $X$, $\mathcal{G}$ over $Y$ and $\mathcal{O}$ denoting the monoidal unit.

Cohomology:........................................... $\mathrm{Hom}_X(\mathcal{O}_X,\mathcal{F})$

Homology:.............................................. $\mathrm{Hom}_X(\mathcal{O}_X,f^!\mathcal{G})$

Cohomology with compact support:....... $\mathrm{Hom}_Y(\mathcal{O}_Y,f_!\mathcal{F})$

Borel-Moore homology:......................... $\mathrm{Hom}_Y(\mathcal{O}_Y,f_!f^!\mathcal{G})$

The defining relations between the 6 functors yield the usual dualities and comparison maps between the four theories. Notice that when f is proper $f_! = f_*$ and the latter pair becomes the first.

• I am confused by the role of O here. Shouldn't the first two rows be just f_*F and f_*f^! G ? While the last become f_!F and f_! f^! G? – Matthias Volkov May 6 '16 at 15:41
• @Mattias Volkov If $Y$ is affine, then your definitions agree with mine (after taking the associated sheaf). In the general case you give a sheafified version of co/homology while my definitions provide abelian groups, in fact modules over $\Gamma(X, \mathcal{O}_X)$ for the first two and over $\Gamma(Y, \mathcal{O}_Y)$ for the remaining ones. – Leo Alonso May 6 '16 at 15:50
• I understand that, but I don't see Y playing any role in your definition. For example, when f is the identity, don't your definitions of "cohomology with compact support" and "cohomology" with coefficients in F coincide? That just seems wrong to me. That's why I thought you should replace the definition with my comment above, which gives the correct answer when Y is a point. [also, in your comment I think you want Y a point (or maybe contractible) because in general affine varieties have non-trivial cohomology.] – Matthias Volkov May 6 '16 at 17:35
• The philosophy behind these definitions is that of relative objects. They express invariants "along the fibers". For an absolute cohomology, use a contractible base or the absolute base. The 6 operations formalism is geared towards a relative version of co/homology in the spirit of Grothendieck. – Leo Alonso May 6 '16 at 22:14
• It's missing a $f_{*}$ in the first one if you want something relative. – user40276 Sep 28 '16 at 6:57