How duality follows from a six functor formalism

Question

For the sake of this question, we'll model a six functor formalism in the following way. Let $\mathsf{C}$ be a category of spaces (be it the category of schemes, or topological spaces) and consider a triangulated closed symmetric monoidal category $\mathsf{D}(X)$, with identity $\mathscr{O}_X$, for each $X\in\mathsf{C}$.

Given a morphism $f:X\to Y$ in $\mathsf{C}$, we have functors $f_*,f_!:\mathsf{D}(X)\leftrightarrows \mathsf{D}(Y):f^*,f^!$ such that $f^*\dashv f_*$, $f_!\dashv f^!$, and $f^*$ is strong symmetric monoidal.

If we also suppose that $\mathsf{C}$ has a final object $S$, it is natural to define cohomology and cohomology with compact support to be $$H^i(X,M):=\hom_{\mathsf{D}(S)}(\mathscr{O}_S,p_*M[i])\qquad\text{and}\qquad H_c^i(X,M):=\hom_{\mathsf{D}(S)}(\mathscr{O}_S,p_!M[i]),$$ where $p:X\to S$ is the natural morphism. If $\mathsf{C}$ is the category of ringed spaces (or even of ringed sites, I think), this coincides with the usual definitions.

I wonder then how could we obtain some sort of duality isomorphism similar to Poincaré and Serre duality. Perhaps we could begin with $$\operatorname{Ext}^i(M,p^!\mathscr{O}_S)=\hom_{\mathsf{D}(X)}(M,p^!\mathscr{O}_S[i])=\hom_{\mathsf{D}(S)}(p_!M[-i],\mathscr{O}_S)$$ and the latter should be something like $H^{-i}_c(X,M)^\vee$, but I'm not sure how to make this precise.

Answer 1

Your description of the six functors does not mention any relations between the $!$-functors and the $*$-functors or the tensor product, which is where these dualities are hiding.

Poincaré duality is a relation between the $*$-functors and the $!$-functors. Typically there are canonical isomorphisms $f_!=f_*$ when $f$ is proper and $f^!=f^*[d]$ when $f$ is nice (e.g. smooth) of relative dimension $d$. Depending on how the $!$-functors are constructed, one of these isomorphisms will hold by definition but the other one will be nontrivial to prove. When $f:X\to S$ is nice, we get isomorphisms (where $1_S$ is the unit object in $D(S)$) $$ H^n(X,M) := \mathrm{hom}(1_S, f_*f^*(M)[n]) = \mathrm{hom}(1_S,f_*f^!(M)[n-d])=:H_{d-n}^\mathrm{BM}(X,M) $$ and $$ H^n_c(X,M):= \mathrm{hom}(1_S, f_!f^*(M)[n]) = \mathrm{hom}(1_S,f_!f^!(M)[n-d])=:H_{d-n}(X,M), $$ On the other hand, there are vertical identifications when $f$ is proper.

Serre duality uses a relation between the $!$-functors and the tensor product. Namely, for $f:X\to S$, the functor $f_!$ is $D(S)$-linear; in particular we have the projection formula $f_!(A\otimes f^*(B))= f_!(A)\otimes B$. By adjunction this implies the isomorphism $$ f_*\mathrm{Hom}(A, f^!(B)) = \mathrm{Hom}(f_!(A),B). $$ One gets Serre duality when $f$ is proper (so $f_!=f_*$) and $B=1_S$ (so $f^!(B)$ is the "dualizing sheaf"). Combining the projection formula with Poincaré duality we can also deduce that $f_*$ preserves $\otimes$-dualizable objects when $f$ is both proper and nice, which gives a perfect pairing. But the usual formulation of Serre duality in terms of cohomology is specific to the derived category of quasi-coherent sheaves when $S=\mathrm{Spec}(k)$ is a field (note that in the quasi-coherent case the adjunction $(f_!,f^!)$ only exists when $f$ is proper).

There are further duality statements, such as Verdier duality, which says that $\mathrm{Hom}(-,f^!(1_S))$ is an equivalence of a certain subcategory $D_c(X)\subset D(X)$ of "constructible"/"coherent" objects with its opposite. But unlike the previous two, this statement does not directly follow from the various relations between the six functors.

Answer 2

@MarcHoyois It would help to have a further comparison between $\mathrm{hom}(1_S, f_!f^!M[m])$ and a naive definition of homology such as $H^{m}(X, M)^\lor$ in the case of field coefficients.