This is the fifth in a series of posts on Verdier duality, started here. In this post, I will describe the proof of the duality theorem, which itself states the existence of an adjoint to the derived version of the lower shriek functor $f_!$. This might not sound too exciting at first, but we will see that in fact, the dualizing functor will be computable in the important special case of a manifold, and Poincaré duality will fall out quickly. Moreover, the flexible interpretation of sheaf cohomology will allow other duality theorems (such as Alexander duality) to be derived very efficiently from the general formalism.

I will try to explain some of this story (namely, that using sheaf cohomology and Verdier duality one can re-derive much of the classical theory of homology and cohomology) next time. First, though, it will be good to prove the result.

1. Duality

We can now enunciate the result we shall prove in full generality.

Theorem 1 (Verdier duality) Let ${f: X \rightarrow Y}$ be a continuous map of locally compact spaces of finite dimension, and let ${k}$ be a noetherian ring. Then ${\mathbf{R} f_!: \mathbf{D}^+(X, k) \rightarrow \mathbf{D}^+(Y, k)}$ admits a right adjoint ${f^!}$. In fact, we have an isomorphism in ${\mathbf{D}^+(k)}$

$\displaystyle \mathbf{R}\mathrm{Hom}( \mathbf{R} f_! \mathcal{F}^\bullet , \mathcal{G}^\bullet) \simeq \mathbf{R}\mathrm{Hom}( \mathcal{F}^\bullet, f^! \mathcal{G}^\bullet)$

when ${\mathcal{F}^\bullet \in \mathbf{D}^+(X, k), \mathcal{G}^\bullet \in \mathbf{D}^+(Y, k)}$.

Here ${\mathbf{R}\mathrm{Hom}}$ is defined as follows. Recall that given chain complexes ${A^\bullet, B^\bullet}$ of sheaves, one may define a chain complex of ${k}$-modules ${\hom^\bullet(A^\bullet, B^\bullet)}$; the elements in degree ${n}$ are given by the product ${\prod_m \hom(A^m, B^{m+n})}$, and the differential sends a collection of maps ${\left\{f_m: A^m \rightarrow B^{m+n}\right\}}$ to ${ df_m + (-1)^{n+1} f_{m+1}d: A^m \rightarrow B^{m+n+1}}$. Then ${\mathbf{R}\mathrm{Hom}}$ is the derived functor of ${\hom^\bullet}$, and lives in the derived category ${\mathbf{D}^+(k)}$ if ${A^\bullet, B^\bullet \in \mathbf{D}^+(X, k)}$. Since the cohomology in degree zero is given by ${\hom_{\mathbf{D}^+(X, k)}(A^\bullet, B^\bullet)}$, we see that the last statement of Verdier duality implies the adjointness relation.

You might wonder why Verdier duality is so much more complex than classical Poincaré duality. I don’t know a complete answer to this, but the impression I get is the following: Verdier duality is Poincaré duality when the upper shriek is a simple functor. When $X$ is a manifold and $k$ is a field, the upper shriek sends the complex consisting of $k$ in degree zero (which generates the derived category of $k$-vector spaces) to a complex of sheaves cohomologically concentrated in one degree. Maps in the derived category between complexes cohomologically concentrated in one degree are easy to describe using classical constructions (namely, they are $\mathrm{Ext}$ groups), and so Poincaré duality does not require derived categories to state.

In general, however, the upper shriek functor is much more complicated, and the resulting complexes don’t have to be cohomologically concentrated in one degree. So derived categories seem to be indispensable. It seems that this is the same reason that Grothendieck duality (which is formally analogous to Verdier duality) is so much more complex than Serre duality (which is formally analogous to Poincaré duality): in the case of a smooth variety, the analog of the upper shriek functor for coherent sheaves always sends the ground field to a complex cohomologically concentrated in degree zero!

(In the above half-explanation, you might also object that Poincare duality as traditionally stated (say for a field) does not involve $\mathrm{Ext}$ groups; the explanation is analogous to the translation between the various formulations of Serre duality, and I will give it next time.)

2. Representability

Ultimately, the existence of an adjoint to a functor ${F: \mathcal{C} \rightarrow \mathcal{D}}$ is equivalent to representability of the functor ${C \mapsto \hom_{\mathcal{D}}(FC, D)}$ for each ${D \in \mathcal{D}}$. Following Gelfand and Manin, we thus start by proving a simple representability lemma.

Lemma 2 Let ${X}$ be a space. An additive functor ${F: \mathbf{Sh}(X, k) \rightarrow k\mbox{--}\mathrm{mod}^{op}}$ is representable if and only if it sends colimits to limits.

The point of this result is that, while the ordinary functor ${f_!}$ is not (generally) a left adjoint, something very close to it is. That something very close will be in fact isomorphic in the derived category. We shall see this below.

Proof: One direction is always true (in any category). Suppose conversely that ${F}$ is representable. The strategy is that ${\mathbf{Sh}(X, k)}$ has a lot of generators: namely, for each open set ${U \subset X}$, take ${k_U = j_!(k)}$ where ${j: U \rightarrow X}$ is the inclusion. We can define a sheaf ${\mathcal{F} \in \mathbf{Sh}(X, k)}$ via

$\displaystyle \mathcal{F}(U) = F(k_U).$

Since the ${k_U}$ have canonical imbedding maps (if ${U \subset U',}$ there is a map ${k_U \rightarrow k_{U'}}$), it is clear that ${\mathcal{F}}$ is a presheaf. ${\mathcal{F}}$ is in fact a sheaf, though. To see this, let ${\left\{U_\alpha\right\}}$ be an open covering of ${U}$; then there is an exact sequence of sheaves

$\displaystyle \prod_{\alpha, \beta} k_{U_\alpha \cap U_\beta} \rightarrow \prod_\alpha k_{U_\alpha} \rightarrow k_U \rightarrow 0,$

which means that there is an exact sequence

$\displaystyle 0 \rightarrow F(k_U ) \rightarrow \prod F(k_{U_\alpha}) \rightarrow \prod_{\alpha, \beta} F(k_{U_\alpha \cap U_\beta}).$

This means that ${\mathcal{F}}$ is a sheaf. ${\mathcal{F}}$ is a promising candidate for a representing object, because we know that

$\displaystyle \hom(k_U, \mathcal{F}) \simeq \mathcal{F}(U) = F(k_U).$

Now, we need to define a distinguished element of ${F(\mathcal{F})}$ and show that it is universal. More generally, we can define a natural transformation ${\hom(\cdot, \mathcal{F}) \rightarrow F(\cdot)}$. This we can do because any ${\mathcal{G} \in \mathbf{Sh}(X, k)}$ is canonically a colimit of sheaves ${k_U}$.

Namely, form the category whose objects are pairs ${(U, s)}$ where ${U \subset X}$ is open and ${s \in \mathcal{G}(U)}$ and whose morphisms come from inclusions ${(V, s') \rightarrow (U, s)}$ where ${V \subset U}$ and ${s' = s|_V}$. For each such pair define a map ${k_U \rightarrow \mathcal{G}}$ by the section ${s}$. It is easy to see that this gives a representation of ${\mathcal{G}}$ functorially as a colimit of sheaves of the form ${k_U}$.

(This is a similar observation as the well-known fact that a presheaf on any small category is canonically colimit of representable presheaves.) The natural isomorphism ${\hom(k_U, \mathcal{F}) \simeq F(k_U)}$ now extends to a natural transformation ${\hom(\mathcal{G}, \mathcal{F}) \simeq F(\mathcal{G})}$, which is an isomorphism. Indeed, it is an isomorphism when ${\mathcal{G} = k_U}$, and both functors above commute with colimits.

3. A near-adjoint to ${f_!}$

In an ideal world, the functor ${f_!}$ would already have an adjoint. That is, for each ${\mathcal{G} \in \mathbf{Sh}(Y, k)}$, the functor

$\displaystyle \mathcal{F} \mapsto \hom_{\mathbf{Sh}(Y, k)}(f_! \mathcal{F}, \mathcal{G})$

would be representable. This is not always the case, however: ${f_!}$ is not (probably) an exact functor, and it need not preserve colimits. Nonetheless, a slight variant of the above functor is representable:

Proposition 3 If ${\mathcal{M}}$ is a soft, flat sheaf in ${\mathbf{Sh}(X, k)}$, then the functor ${\mathcal{F} \mapsto f_!(\mathcal{F} \otimes_k \mathcal{M})}$ commutes with colimits. In particular, the functor ${\mathcal{F} \mapsto \hom_{\mathbf{Sh}(Y, k)}(f_! \mathcal{F}, \mathcal{G}) }$ is representable for any ${\mathcal{G} \in \mathbf{Sh}(Y, k)}$.

Proof: Indeed, we know that ${f_!}$ commutes with filtered colimits (because compactly supported cohomology does!) and in particular with arbitrary direct sums. As a result, it suffices to show that ${\mathcal{F} \mapsto f_!(\mathcal{F} \otimes_k \mathcal{M})}$ is an exact functor. If ${0 \rightarrow \mathcal{F}' \rightarrow \mathcal{F} \rightarrow \mathcal{F}'' \rightarrow 0}$ is a short exact sequence in ${\mathbf{Sh}(X, k)}$, then so is ${0 \rightarrow \mathcal{F}' \otimes_k \mathcal{M} \rightarrow \mathcal{F} \otimes_k \mathcal{M} \rightarrow \mathcal{F}'' \otimes_k \mathcal{M} \rightarrow 0}$ by flatness. Moreover, by \cref{tensorsoft} the first term is soft, so the push-forward sequence

$\displaystyle 0 \rightarrow f_!(\mathcal{F}' \otimes_k \mathcal{M}) \rightarrow f_!(\mathcal{F} \otimes_k \mathcal{M} ) \rightarrow f_!(\mathcal{F}'' \otimes_k \mathcal{M}) \rightarrow 0$

is exact too. The representability criterion now completes the proof. It follows that given ${\mathcal{M}}$ (soft, flat) and ${\mathcal{G} \in \mathbf{Sh}(Y, k)}$ as above, there is a sheaf ${f_\# (\mathcal{M}, \mathcal{G}) \in \mathbf{Sh}(X, k)}$ such that

$\displaystyle \hom_{\mathbf{Sh}(X, k)}(\mathcal{F}, f_\#(\mathcal{M}, \mathcal{G})) \simeq \hom_{\mathbf{Sh}(Y, k)}(f_!(\mathcal{F} \otimes_k \mathcal{M}), \mathcal{G}).$

This is clearly functorial in ${\mathcal{G}}$ and contravariantly in ${\mathcal{M}}$. We shall use this functor ${f_\#}$ to construct the adjoint ${f^!}$ in Verdier duality when ${\mathcal{M}}$ is replaced by a complex consisting of soft, flat sheaves.

4. Proof of Verdier duality

We are now ready to prove Verdier duality. The strategy will be to choose a soft, flat, and bounded resolution ${\mathcal{L}^\bullet}$ of the constant sheaf ${k}$, so a quasi-isomorphism ${k \rightarrow \mathcal{L}^\bullet}$.

We have already seen that we can do most of this; however, we should check that we can choose ${\mathcal{L}^\bullet}$ to be bounded. To do this, we truncate ${\mathcal{L}^\bullet}$ after the ${n}$th stage, where ${n = \dim X}$; the resulting complex will remain soft by \cref{softres}. Namely, we consider the complex

$\displaystyle \tau_{\leq n+1}\mathcal{L}^\bullet: 0 \rightarrow \mathcal{L}^0 \rightarrow \mathcal{L}^1 \rightarrow \dots \rightarrow \mathcal{L}^n \rightarrow \mathrm{im}(\mathcal{L}^n \rightarrow \mathcal{L}^{n+1}) \rightarrow 0.$

Since ${\mathcal{L}^1, \dots, \mathcal{L}^n}$ are soft, so is the final term as ${n = \dim X}$. The final term is also flat because of the stalkwise split nature of the resolution ${\mathcal{L}^\bullet}$ (at least if it was constructed using a Godement resolution). Then ${\mathcal{F}^\bullet}$ and ${\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet}$ will be isomorphic functors on the level of the derived categories, but the latter will be much better behaved; for instance, it will have soft terms.

In other words, we are going to obtain a functorial “soft (and thus ${f_!}$-acyclic) replacement” for a given object in ${\mathbf{D}^+(X, k)}$. Fix a complex ${\mathcal{G}^\bullet \in \mathbf{D}^+(Y, k)}$. We need to show that the functor ${\mathcal{F}^\bullet \mapsto \hom_{\mathbf{D}^+(Y, k)}(\mathbf{R} f_!(\mathcal{F}^\bullet), \mathcal{G}^\bullet)}$ is representable. However, there is a canonical isomorphism (in the derived category, or a quasi-isomorphism on the level of complexes)

$\displaystyle \mathcal{F}^\bullet \simeq \mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet.$

This works in the derived category (we do not need to take the derived tensor product) as ${\mathcal{L}^\bullet}$ is flat and bounded. So, alternatively, we may show that the functor ${\mathcal{F}^\bullet \mapsto \hom_{\mathbf{D}^+(Y, k)}(\mathbf{R} f_!(\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet), \mathcal{G}^\bullet)}$ is representable. We shall in fact show that there is a complex ${\mathcal{K}^\bullet \in \mathbf{D}^+(X, k)}$ such that there is a functorial isomorphism

$\displaystyle \mathbf{R}\mathrm{Hom}(\mathbf{R} f_!(\mathcal{F}^\bullet), \mathcal{G}^\bullet) \simeq \mathbf{R}\mathrm{Hom}(\mathbf{R} f_!(\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet), \mathcal{G}^\bullet) \simeq \mathbf{R}\mathrm{Hom}(\mathcal{F}^\bullet, \mathcal{K}^\bullet). \ \ \ \ \ (1)$

This is what we want for the stronger form of Verdier duality anyway. If we wade through this notation, we notice one thing: ${\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet}$ is already ${f_!}$-acyclic; in particular,

$\displaystyle \mathbf{R} f_!(\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet) \simeq f_!(\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet)$

where ${f_!}$ is applied pointwise. Moreover, we can assume that ${\mathcal{G}^\bullet}$ is a complex of injectives, and certainly we can (and will) try to choose ${\mathcal{K}^\bullet}$ to consist of injectives. In this case, we are just looking for a quasi-isomorphism

$\displaystyle \hom^\bullet (f_! (\mathcal{F}^\bullet \otimes_k \mathcal{L}^\bullet), \mathcal{G}^\bullet) \stackrel{qis}{\simeq} \hom^\bullet(\mathcal{F}^\bullet, \mathcal{K}^\bullet)$

valid for any complex ${\mathcal{F}^\bullet}$. However, we know that

If we consider the double complex given by ${C^{rs} = f_\#(\mathcal{L}^{-r}, \mathcal{G}^s)}$ (with the boundary maps being those induced by ${\mathcal{L}, \mathcal{G}}$; remember that ${f_\#}$ is contravariant in the first variable) and let ${\mathcal{K}^\bullet}$ be the associated chain complex with ${\mathcal{K}^t = \bigoplus_{r + s =t}f_\#(\mathcal{L}^{-r}, \mathcal{G}^s)}$, then it follows that there is an isomorphism

$\displaystyle \hom^n(f_!(\mathcal{F}^\bullet \otimes \mathcal{L}^\bullet), \mathcal{G}^\bullet) \simeq \hom^n(\mathcal{F}^\bullet, \mathcal{K}^\bullet).$

In fact, there is an isomorphism of complexes

$\displaystyle \hom^\bullet( f_!(\mathcal{F}^\bullet \otimes \mathcal{L}^\bullet), \mathcal{G}^\bullet) \simeq \hom^\bullet(\mathcal{F}^\bullet, \mathcal{K}^\bullet).$

This follows from checking through the signs of the differential. This will prove (1) if we check that ${\mathcal{K}^\bullet}$ is a bounded-below complex of injectives. It is bounded below from the definition (as ${\mathcal{L}^\bullet}$ is bounded in both directions). To see that it is injective, we recall that we had chosen ${\mathcal{G}^\bullet}$ to be a complex of injectives, make the observation:

Lemma 4 ${f_\#(\mathcal{M}, \mathcal{G})}$ is injective whenever ${\mathcal{M} \in \mathbf{Sh}(X, k)}$ is a soft, flat sheaf and ${\mathcal{G} \in \mathbf{Sh}(Y, k)}$ is injective .

Proof: Recall that ${f_\#(\mathcal{M}, \mathcal{G})}$ is the object representing the functor

$\displaystyle \mathcal{F} \mapsto \hom_{\mathbf{Sh}(Y, k)}(f_!(\mathcal{F} \otimes \mathcal{M}) , \mathcal{G}).$

To say that it is injective is to say that mapping into it is an exact functor, or simply that it is a right exact functor. Let ${0 \rightarrow \mathcal{F}' \rightarrow \mathcal{F}}$ be an exact sequence. Then

$\displaystyle 0 \rightarrow f_!(\mathcal{F}' \otimes \mathcal{M}) \rightarrow f_!(\mathcal{F} \otimes \mathcal{M})$

is exact too, so injectivity of ${\mathcal{G}}$ gives that

$\displaystyle \hom_{\mathbf{Sh}(Y, k)}(f_!(\mathcal{F} \otimes \mathcal{M}), \mathcal{G}) \rightarrow \hom_{\mathbf{Sh}(Y, k)}(f_!(\mathcal{F}' \otimes \mathcal{M}), \mathcal{G}) \rightarrow 0$

is also exact. This statement is the meaning of injectivity.

It is now clear how we may define the functor ${f^!: \mathbf{D}^+(Y, k) \rightarrow \mathbf{D}^+(X, k)}$. Given a bounded-below complex ${\mathcal{G}^\bullet \in \mathbf{D}^+(Y, k)}$, we start by replacing it with a complex of injectives, and so just assume that it consists of injectives without loss of generality. We then form the complex ${\mathcal{K}^\bullet}$ of sheaves on ${X}$ such that ${\mathcal{K}^t = \bigoplus_{r + s = t} f_\#(\mathcal{L}^{-r}, \mathcal{G}^s)}$, where ${\mathcal{L}^\bullet}$ is a fixed soft resolution of the constant sheaf. Then setting ${f^!\mathcal{G}^\bullet = \mathcal{K}^\bullet}$ finishes the proof, by (1); we have functoriality built in.