This is the third in a series of posts started here (in particular, the notation is kept from there) intended to cover the basics of Verdier duality. Here, I will discuss the lower shriek functors needed even to state Verdier duality (in the most general form, at least); as we will see, the class of soft sheaves will be acyclic with respect to this functor. To see this, though, we shall need to prove some general facts on how push-forward behave with respect to base change, which are themselves of independent interest.

1. The ${f_!}$ functors

Let ${f: X \rightarrow Y}$ be a map of spaces. We have defined the functor

$\displaystyle f_! : \mathbf{Sh}(X) \rightarrow \mathbf{Sh}(Y)$

earlier, such that ${f_!(U) }$ consists of the sections of ${\mathcal{F}(f^{-1}(U))}$ whose support is proper over ${U}$ ; ${f_!\mathcal{F} }$ is always a subsheaf of ${f_*\mathcal{F}}$, equal to it if ${f}$ is proper. When ${Y}$ is a point, we get the functor

$\displaystyle \mathcal{F} \mapsto \Gamma_c(X, \mathcal{F}) = \left\{\text{global sections with proper support}\right\}.$

One can check that ${f_! \mathcal{F}}$ is in fact a sheaf. The observation here is that a map ${A \rightarrow B}$ of topological spaces is proper if and only if there is an open cover ${\left\{B_i\right\}}$ of ${B}$ such that ${A \times_B B_i \rightarrow B_i}$ is proper for each ${i}$. Now ${f_!}$ is a left-exact functor, as one easily sees. We now want to show that the class of soft sheaves is acyclic with respect to ${f_!}$, and in particular so that one may use soft resolutions to compute the derived functors. To do this, we shall prove a general “base change” theorem that will compute the stalk of ${f_! \mathcal{F}}$.

Definition 1 As usual, ${R^i f_!}$ denotes the ${i}$th right derived functor of ${f_!}$. There is also a total derived functor on the derived categories ${\mathbf{R} f_!: \mathbf{D}^+(X) \rightarrow \mathbf{D}^+(Y)}$ (and similarly ${\mathbf{D}^+(X, k) \rightarrow \mathbf{D}^+(Y, k)}$).

2. Base change theorems

To understand the functor ${f_!}$ and its derived functors ${R^i f_!}$, we will need to determine their stalks. We start by describing the situation for a proper map.

Theorem 2 (Proper base change) Let ${f: X \rightarrow Y}$ be a proper map of locally compact spaces. If ${\mathcal{F} \in \mathbf{Sh}(X)}$, then there is a functorial isomorphism

$\displaystyle (R^i f_* \mathcal{F})_y \simeq H^i(X_y, \mathcal{F}_y)$

for each ${y \in Y}$.

Here ${\mathcal{F}_y}$ is the restriction of ${\mathcal{F}}$ to ${X_y}$, which in turn is the fiber over ${y}$.

Proof: Note that both ${(R^i f_* \mathcal{F})_y, H^i(X_y, \mathcal{F}_y)}$ are ${\delta}$-functors from ${\mathbf{Sh}(X)}$ to the category of abelian groups. Let us first define the natural transformation. Indeed, we know that ${R^i f_* \mathcal{F}}$ is the sheaf on ${Y}$ associated to the presheaf

$\displaystyle U \mapsto H^i(f^{-1}(U), \mathcal{F}).$

As a result, the stalk at ${y}$ is the direct limit $$\varinjlim_U H^i(f^{-1}(U), \mathcal{F}).$$ Each ${H^i(f^{-1}(U), \mathcal{F})}$ maps naturally to ${H^i(X_y, \mathcal{F}_y)}$, so we get the natural map (of ${\delta}$-functors, even). Let us show that it is an isomorphism when ${i=0}$. This equates to saying that the map

$\displaystyle \varinjlim \mathcal{F}(f^{-1}(U)) \rightarrow \Gamma(X_y, \mathcal{F}|_{X_y})$

is an isomorphism.

But since ${f}$ is a closed map, and ${X_y}$ is compact, it follows that the sets ${f^{-1}(U)}$ for ${U}$ open and containing ${y}$ form a cofinal family in the set of open sets containing ${X_y}$. The claim is now clear because a section over a compact set automatically extends to a small neighborhood.

Finally, we need to show that both functors are effaceable in positive dimensions. For ${(R^i f_* \cdot)_y}$, this is immediate. For the other functor, let us show that if ${\mathcal{F}}$ is soft, then ${H^i(X_y, \mathcal{F}_y) =0}$ for ${i> 0}$. Since any sheaf can be imbedded in a soft (e.g. flabby) sheaf, this will be enough. But this in turn is clear because ${\mathcal{F}_y}$ is itself then soft:

Lemma 3 The restriction of a soft sheaf to a locally closed subspace is soft.

Proof: Indeed, this follows because if ${\mathcal{F} \in \mathbf{Sh}(X)}$ is soft and ${Z \subset X}$, then any compact subset of ${Z}$ is a compact subset of ${X}$. So a section of ${\mathcal{F}}$ over ${Z}$ can be extended all the way over ${X}$, and a fortiori over ${Z}$.

Here is the analog for the ${f_!}$ functor:

Theorem 4 Let ${f: X \rightarrow Y}$ be a continuous map of locally compact spaces. If ${\mathcal{F} \in \mathbf{Sh}(X)}$, then there is a functorial isomorphism

$\displaystyle (R^i f_! \mathcal{F})_y \simeq H^i_c(X_y, \mathcal{F}_y)$

for each ${y \in Y}$.

Proof: This is now proved as before. Again, the key point is that a map of ${\delta}$-functors can easily be defined, which is an isomorphism in degree zero, and both functors are effaceable in positive degrees.

But there are some subtleties! For one thing, it is not true that ${R^i f_! \mathcal{F}}$ is the sheaf associated to the presheaf ${U \mapsto H^i_c(U, \mathcal{F})}$. This fails even if ${i =0}$ (take ${f}$ the identity map, for instance). However, by general nonsense, if we can define an isomorphism in degree zero, and show that both functors are effaceable, then we will be done. Both functors are effaceable since ${R^i f_!}$ is effaceable (being a derived functor) while ${\mathcal{F} \mapsto H^i_c(X_y, \mathcal{F}_y)}$ vanishes for soft sheaves.

So we just have to check that ${(f_! \mathcal{F}) _y = \Gamma_c(X_y, \mathcal{F}_y)}$. Now there is a natural ${(f_! \mathcal{F})_y \rightarrow \Gamma_c(X_y, \mathcal{F}_y)}$ that sends a section of ${\mathcal{F}(f^{-1}(U))}$, with support proper over ${U}$, to the restriction to ${X_y}$. This map is injective; for if a section ${s \in \mathcal{F}(f^{-1}(U))}$ with proper support was zero on ${f^{-1}(y)}$, then the image of ${\mathrm{supp}( s)}$ in ${U}$ would not contain ${y}$. But by properness this image is closed, so we can find a smaller neighborhood ${V}$ containing ${y}$ such that ${\mathrm{supp} s \cap f^{-1}(V) = \emptyset}$. Now we need to check surjectivity.

This is a bit tricky to do directly, but fortunately a trick helps out here. We know that ${(f_! \mathcal{F})_y \rightarrow \Gamma_c(X_y, \mathcal{F})}$ is surjective when ${\mathcal{F}}$ is soft: this is easy to see (for an element of ${\Gamma_c(X_y, \mathcal{F})}$ can be extended to all of ${X}$ with compact, and certainly proper support). Now given ${\mathcal{F}}$, we find an exact sequence

$\displaystyle 0 \rightarrow \mathcal{F} \rightarrow \mathcal{I} \rightarrow \mathcal{J}$

with ${\mathcal{I}, \mathcal{J}}$ soft (e.g. injective). Since the map from the stalk to ${\Gamma_c(X_y, \cdot)}$ is an isomorphism for ${\mathcal{I}, \mathcal{J}}$, and since both functors are obviously left exact, a diagram chase shows that the map is an isomorphism for ${\mathcal{F}}$.

3. Applications

We start by proving a result that provides some content to the phrase “base change.”

Theorem 5 Consider a cartesian diagram of locally compact Hausdorff spaces:

Then there is a natural isomorphism, for any ${\mathcal{F}^\bullet \in \mathbf{D}^+(X)}$,

$\displaystyle f^* \mathbf{R} p_! \mathcal{F}^\bullet \simeq \mathbf{R} p'_! f'^* \mathcal{F}^\bullet .$

Taking cohomology, one sees that for a sheaf ${\mathcal{F}}$, one has natural isomorphisms

$\displaystyle f^* R^i p_! \mathcal{F} \simeq R^i p'_! f'^* \mathcal{F}.$

Proof:One may define this map by the universal property, on the level of complexes. Namely, let ${\mathcal{G}^\bullet}$ be a complex of sheaves. We will define a map

$\displaystyle f^* p_! \mathcal{G}^\bullet \rightarrow p'_! f'^* \mathcal{G}^\bullet.$

To do this, we may as well assume ${\mathcal{G}^\bullet}$ is a single sheaf ${\mathcal{G}}$ (by naturality). So we are reduced to defining a map

$\displaystyle f^* p_! \mathcal{G}\rightarrow p'_! f'^* \mathcal{G}, \quad \mathcal{G}\in \mathbf{Sh}(X).$

This is equivalent to defining a map ${p_! \mathcal{G}\rightarrow f_* p'_! f^* \mathcal{G}}$. Given a section of ${p_! \mathcal{G}}$ over an open set ${U}$, or equivalently a section ${s}$ of ${\mathcal{G}(p^{-1}(U))}$ with proper support over ${U}$, we can consider this as a section of ${f^* \mathcal{G}}$ over ${f'^{-1}(p^{-1}(U)) = p'^{-1}(f^{-1}(U))}$ with proper support over ${U}$, and thus over ${f^{-1}(U)}$. This is equivalently a section of ${p'_! f^*\mathcal{G}}$ over ${f^{-1}(U)}$, or a section of ${f_* p'_! f^* \mathcal{G}}$ over ${U}$.

So we can get the base change morphism, which is clearly natural. From here it is easy to see that it can be defined even in the derived category. To check that it is an isomorphism, we reduce by general facts on “way-out” functors (proved in Hartshorne’s Residues and Duality, for instance) to showing that the map is an isomorphism for a single sheaf ${\mathcal{F}}$, or equivalently that

$\displaystyle f^* R^i p_! \mathcal{F} \simeq R^i p'_! f'^* \mathcal{F}.$

But now this follows by taking the stalks at some ${s' \in S}$; on the left, we get ${H^i_c(p^{-1}(f(s)), \mathcal{F})}$, and on the right we get ${H^i_c(p'^{-1}(s'), f'*\mathcal{F})}$, which are both the same since these are isomorphic spaces. (By abuse of notation we have written ${\mathcal{F}}$ for the restrictions to various subspaces.)

As another example of these base change theorems, we prove the promised result that soft sheaves are acyclic with respect to the lower shriek.

Proposition 6 Let ${0 \rightarrow \mathcal{F}' \rightarrow \mathcal{F} \rightarrow \mathcal{F}'' \rightarrow 0}$ be an exact sequence of sheaves on ${X}$. Suppose ${\mathcal{F}'}$ is soft. Then the sequence ${0 \rightarrow f_!\mathcal{F}' \rightarrow f_! \mathcal{F} \rightarrow f_!\mathcal{F}'' \rightarrow 0}$ is also exact, and ${\mathbf{R} f_! \mathcal{F}}$ is cohomologically concentrated in degree zero (or, what is the same thing, ${R^i f_! \mathcal{F}}$ vanish for ${i > 0}$).

Proof: It follows that the sequence ${0 \rightarrow f_!\mathcal{F}' \rightarrow f_! \mathcal{F} \rightarrow f_!\mathcal{F}'' \rightarrow 0}$ is exact if ${R^1 f_! \mathcal{F}' = 0}$. But more generally ${R^i f_! \mathcal{F}' = 0}$ because taking stalks at each ${y \in Y}$ gives ${H^i_c(X_y, \mathcal{F}') = 0}$. Thus, soft resolutions will suffice to compute ${\mathbf{R} f_!}$, which will be very convenient.

As another example of base change, consider an open immersion ${j: U \rightarrow X}$. By looking at stalks (and defining a natural map), we find that ${j_! \mathcal{F}}$ for ${\mathcal{F} \in \mathbf{Sh}(U)}$ is just “extension by zero.”

4. The Leray spectral sequence for ${\mathbf{R} f_!}$

In fact, ${f_!}$ even preserves soft sheaves, which leads to a Leray spectral sequence for ${\mathbf{R} f_!}$.

Proposition 7 If ${f: X \rightarrow Y}$ is continuous and ${\mathcal{F} \in \mathbf{Sh}(X)}$ soft, then ${f_! \mathcal{F}}$ is soft too.

Proof: Indeed, let ${Z \subset Y}$ be a compact set, and suppose ${s \in \Gamma(Z, f_! \mathcal{F})}$ is a section, so we know that ${s}$ extends to a small neighborhood ${U}$ of ${Z}$; call the extension ${\widetilde{s}}$. Then this extension ${\widetilde{s}}$ of ${s}$ becomes a section of ${f^{-1}(U)}$ with proper support.

Restricting ${U}$ to a compact neighborhood of ${Z}$, we get a compactly supported section of ${\mathcal{F}}$, which we can extend to all of ${X}$ so as to have compact support (and thus get a global section of ${f_! \mathcal{F}}$).

Corollary 8 (Leray spectral sequence) Given maps ${f: X \rightarrow Y, g: Y \rightarrow Z}$ of locally compact Hausdorff spaces, there is a natural isomorphism ${\mathbf{R} (g \circ f)_! \simeq \mathbf{R} g_! \circ \mathbf{R} f_!}$, and a spectral sequence for any ${\mathcal{F} \in \mathbf{Sh}(X)}$,

$\displaystyle R^i g_! R^j f_! \mathcal{F} \implies R^{i+j}(g \circ f)_! \mathcal{F}.$

Proof: This is now clear, because ${f_!}$ maps injective sheaves (which are flasque, hence soft) to soft sheaves, which are ${g_!}$-acyclic, and we can apply the general theorem.