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- \documentclass{../../notes}
-
- \newcommand{\com}[1]{#1^{\text{\scalebox{0.7}{\textbullet}}}}
- \newcommand{\K}{\mathcal{K}}
- \renewcommand{\lim}{\varprojlim}
- \newcommand{\colim}[1]{\underset{#1}{\operatorname{colim}\;}}
-
- \newcommand{\spec}{\operatorname{Spec }}
-
- \newcommand{\sh}[1]{\mathcal{A}b(#1)}
- \newcommand{\supp}[1]{\operatorname{supp}(#1)}
-
- \title{Lower shriek}
- \author{Christian Merten}
-
- \begin{document}
-
- \section{Preliminaries}
-
- These notes mostly follow \cite{mathew}. Some ideas are taken
- from \cite{gelfand} and \cite{kashiwara}.
-
- In the following, a topological space $X$ is always assumed to be locally compact and Hausdorff.
- Denote by $\sh{X}$ the category
- of sheaves of abelian groups on $X$. Furthermore, denote by
- $\mathrm{D}(X)$ (respectively $\mathrm{D}^{+}(X)$) the (bounded below) derived category of $\sh{X}$.
-
- \begin{definition}[Lower Shriek]
- Let $f\colon X \to Y$ be a continuous map of spaces.
- For $\mathcal{F} \in \sh{X}$
- and $U \subseteq Y$ open, let
- \[
- f_{!}(\mathcal{F})(U) = \{ s \in \mathcal{F}(f^{-1}(U)) \colon \supp{s} \xrightarrow{f} U \text{ proper}\}
- .\]
- \end{definition}
-
- \begin{bem}[Support]
- For $\mathcal{F} \in \sh{X}$, $U \subseteq X$ open and a section $s \in \mathcal{F}(U)$,
- its support $\supp{s}$ is defined as
- \[
- \{ x \in U\colon s_x \neq 0\}
- .\] This set is always closed, as its complement is open.
- \end{bem}
-
- \color{gray}
-
- \begin{lemma}[Lower shriek of sheaf is a sheaf]
- Let $\mathcal{F} \in \sh{X}$ be a sheaf $f\colon X \to Y$ continuous.
- Then $f_{!}\mathcal{F}$ is a sheaf on $Y$.
- \end{lemma}
-
- \begin{proof}
- Clearly, $f_{!}\mathcal{F}$ is a sub-presheaf of the sheaf $f_{*} \mathcal{F}$. To show
- it is a sheaf, we need to verify that gluing sections in $f_{!}\mathcal{F}$ gives again a
- section in $f_{!}\mathcal{F}$.
-
- Let $(U_i)_{i \in I}$ be a family of open sets in $Y$ and $s_i \in (f_{!} \mathcal{F})(U_i)$
- sections. Thus $s_i \in \mathcal{F}(f^{-1}(U_i))$ such that $\supp{s_i} \xrightarrow{f} U_i$
- is proper.
- Gluing yields a unique section $s \in \mathcal{F}(f^{-1}(U))$. We need
- to check that
- \[
- \supp{s} = \bigcup_{i \in I} \supp{s_i} \xlongrightarrow{f} \bigcup_{i \in I} U_i
- \] is proper. For this note that
- $\left(f|_{\supp{s}}\right)^{-1}(U_i) = f^{-1}(U_i) \cap \supp{s} = \supp{s_i}$ and
- being proper is local on the target.
- \end{proof}
-
- \color{black}
-
- \begin{bem}[Lower shriek is left exact]
- Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}''$ be an exact sequence
- in $\sh{X}$ and $f\colon X \to Y$ continuous. Then
- \[
- 0 \to f_{!} \mathcal{F}' \to f_{!}\mathcal{F} \to f_{!}\mathcal{F}''
- \] is exact.
- \end{bem}
-
- \color{gray}
-
- \begin{proof}
- We have the following commutative diagram
- \[
- \begin{tikzcd}
- 0 \arrow{r} & f_{!} \mathcal{F}' \arrow{r} \arrow[hookrightarrow]{d}
- & f_{!} \mathcal{F} \arrow{r} \arrow[hookrightarrow]{d}
- & f_{!} \mathcal{F}'' \arrow[hookrightarrow]{d} \\
- 0 \arrow{r} & f_{*} \mathcal{F}' \arrow{r}
- & f_{*} \mathcal{F} \arrow{r}
- & f_{*} \mathcal{F}''
- \end{tikzcd}
- ,\] where the second row is exact. Thus the claim follows.
- \end{proof}
-
- \color{black}
-
- \begin{bem}[Lower shriek and compact support]
- Let $f\colon X \to \{ *\} $ be the unique continuous map from $X$ to the one point space
- and $\mathcal{F} \in \sh{X}$.
- Then
- \[
- (f_{!}\mathcal{F})(\{*\}) =
- \{ s \in \mathcal{F}(X)\colon \supp{s} \to \{ *\} \text{ proper}\}
- = \{ s \in \mathcal{F}(X)\colon \supp{s} \text{ compact}\}
- .\] Denote this by $\Gamma_c(X, \mathcal{F})$.
- \end{bem}
-
- \section{Derivative of lower shriek}
-
- The goal of this and the following talk is to prove the following theorem
-
- \begin{theorem}[Verdier duality]
- If $X, Y$ are locally compact topological spaces of finite dimension,
- then $\mathrm{R}f_{!}$ admits a right adjoint
- $f^{!}\colon \mathrm{D}^{+}(Y) \to \mathrm{D}(X)$.
- \end{theorem}
-
- To calculate the derivative of $f_{!}$, we need to introduce an adapted class of sheaves.
-
- \begin{definition}
- Let $X$ be space, $\mathcal{F} \in \sh{X}$ and $Z \subseteq X$ a subset. Then
- define
- \[
- \mathcal{F}(Z) = \Gamma(Z, \mathcal{F}) = \Gamma(Z, \mathcal{F}|_{Z})
- \] where $\mathcal{F}|_{Z} = i^{*}\mathcal{F}$ for $i\colon Z \to X$ the canonical inclusion.
- \end{definition}
-
- \begin{bem}[Lemma 1.4 in \cite{mustata}]
- If $\mathcal{F} \in \sh{X}$, $Z_1, Z_2 \subseteq X$ are closed
- and $t_1 \in \mathcal{F}(Z_1)$, $t_2 \in \mathcal{F}(Z_2)$ are given such that
- $t_1|_{Z_1 \cap Z_2} = t_2|_{Z_1 \cap Z_2}$, then
- there exists a unique section $t \in \mathcal{F}(Z_1 \cup Z_2)$ such that
- $t|_{Z_1} = t_1$ and $t|_{Z_2} = t_2$.
- \end{bem}
-
- \color{gray}
-
- \begin{bem}
- If $Z \subseteq X$ is a subset and $i\colon Z \to X$ the canonical inclusion, then
- \[
- \mathcal{F}(Z)
- =
- \left\{ (s_i, U_i)_{i \in I} \colon U_i \subseteq X \text{ open with } Z \subseteq \bigcup_{i \in I} U_i,
- s_i \in \mathcal{F}(U_i) \text{ with } (s_i)_z = (s_{j})_z \forall i, j \in I, z \in Z \cap U_i \cap U_j\right\} / \sim
- .\]
- where $(U_i, s_i)_{i \in I} \sim (V_j, t_j)_{j \in J}$
- if and only if $(s_i)_z = (t_j)_z$ for all $i \in I$, $j \in J$ and $z \in U_i \cap V_j \cap Z$.
-
- For every open neighbourhood $U$ of $Z$, we have a restriction map
- \[
- \mathcal{F}(U) \to \mathcal{F}(Z), s \mapsto s|_Z \coloneqq [(s, U)]
- .\] This induces a map
- \[
- \colim{Z \subseteq U} \mathcal{F}(U)
- \to \mathcal{F}(Z)
- .\]
- \end{bem}
-
- \begin{lemma}
- Let $X$ be a space and $\mathcal{F} \in \sh{X}$.
- If $Z \subseteq X$ is compact, the natural map
- \[
- \colim{Z \subseteq U} \mathcal{F}(U) \longrightarrow \mathcal{F}(Z)
- \] is an isomorphism.
- \end{lemma}
-
- \begin{proof}
- Injectivity: Let $s \in \mathcal{F}(U)$ such that $s|_Z = 0$. Thus for all $z \in Z$,
- $s_z = 0$ and
- there exists an open neighbourhood
- $z \in U_z \subseteq U$ such that $s|_{U_z} = 0$. Thus $s|_{\bigcup U_z } = 0$. Since
- $Z \subseteq \bigcup_{z \in Z} U_z$, $s$ is zero in the colimit.
-
- Surjectivity: Take $(s_i, U_i)_{i \in I} \in \mathcal{F}(Z)$. Thus
- $Z \subseteq \bigcup_{i \in I} U_i$ and by local compactness, for every $z \in Z$, there
- exists a compact neighbourhood $z \in K_z$ such that $K_z \subseteq U_{i_z}$ for
- some $i_z \in I$. Since $Z$ is compact, finitely many suffice, so we may assume
- $Z \subseteq \bigcup_{i=1}^{n} K_i$ and $K_i \subseteq U_i \subseteq X$.
- We now want to define a section on a neighbourhood of $Z$ that locally agrees with the $s_i$.
-
- By induction, we may assume $n = 2$. By definition, $(s_1)_z = (s_2)_z$ for all $z \in Z \cap U_1 \cap U_2$,
- in particular $s_1|_{U_1 \cap U_2}$ and $s_2|_{U_1 \cap U_2}$ have the same restriction
- to $K_1 \cap K_2$. By the injectivity of the restriction map,
- there exists an open neighbourhood $K_1 \cap K_2 \subseteq V \subseteq U_1 \cap U_2$, such that
- $s_1|_V = s_2|_V$. Since $K_j \setminus V$ is closed in the compact $K_j$, for $j=1,2$
- the subset $K_j \setminus V$ is compact. Since $X$ is Hausdorff, there
- exist open neighbourhoods $K_j \setminus V \subseteq U_j' \subseteq U_j$ such that
- $U_1' \cap U_2' = \emptyset$. Now $s_1|_{U_1'}$, $s_2|_{U_2'}$ and
- $s_1|_V = s_2|_V$ glue to a section $w$ on $U_1' \cup U_2' \cup V \supseteq K_1 \cup K_2 \supseteq Z$
- such that $w|_Z = [(s_i, U_i)_{i \in I}]$.
- \end{proof}
-
- \color{black}
-
- \begin{definition}
- A sheaf $\mathcal{F} \in \sh{X}$ is \emph{soft} if
- $\mathcal{F}(X) \to \mathcal{F}(Z)$ is surjective whenever $Z \subseteq X$ is compact.
- \end{definition}
-
- \begin{bem}
- In \cite{kashiwara} our notion of softness is called \emph{c-soft}.
- For $\sigma$-compact spaces the notions agree according to Exercise II.6 in \cite{kashiwara}.
- \end{bem}
-
- \begin{bem}[Flasque sheaves are soft]
- Recall that a sheaf $\mathcal{F} \in \sh{X}$ is called \emph{flasque}, if
- for every open set $U \subseteq X$, the restriction map
- $\mathcal{F}(X) \to \mathcal{F}(U)$ is surjective. For $Z \subseteq X$ compact,
- we have a commutative diagram:
- \[
- \begin{tikzcd}
- \mathcal{F}(X) \arrow{rr} \arrow[twoheadrightarrow]{dr} & & \mathcal{F}(Z) \\
- & \colim{Z \subseteq U} \mathcal{F}(U) \arrow{ur}{\simeq} &
- \end{tikzcd}
- .\] Thus $\mathcal{F}$ is soft.
- \end{bem}
-
- \begin{bem}[Prop. 2.5.6 in \cite{kashiwara}]
- Let $\mathcal{F} \in \sh{X}$. Then $\mathcal{F}$ is soft if and only if for
- any closed subset $Z \subseteq X$, the restriction
- $\Gamma_c(X, \mathcal{F}) \to \Gamma_c(Z, \mathcal{F}|_{Z})$
- is surjective.
- \end{bem}
-
- \color{gray}
-
- \begin{proof}
- If $K \subseteq X$ is compact, $\Gamma(K, F) = \Gamma_c(K, F|_K)$,
- so the condition is sufficient. Conversely
- assume $\mathcal{F}$ is soft and let $s \in \Gamma_c(Z, \mathcal{F}|_Z)$ with
- compact support $K$. Let $U$ be a relatively compact open neighbourhood of $K$ in $X$.
- Define $\tilde{s} \in \Gamma(\partial U \cup (Z \cap \overline{U}), \mathcal{F})$
- by setting $\tilde{s}_{Z \cap \overline{U}} = s$
- and $\tilde{s}|_{\partial U} = 0$. By softness, this extends to a global section
- $t \in \Gamma(X, \mathcal{F})$. Since $t = 0$ on a neighbourhood of $\partial U$,
- we may assume $t$ is supported by $\overline{U}$.
- \end{proof}
-
- \color{black}
-
- \begin{bsp}
- Let $M$ be a smooth manifold and let $f \in \mathcal{C}^{\infty}(K)$ be a
- section over a compact set $K$, i.e. a smooth function defined
- on some neighbourhood $U$ of $K$. Thus by using a partition of unity,
- we can extend $f$ to a global smooth function $\tilde{f} \in \mathcal{C}^{\infty}(M)$
- such that $\tilde{f}|_{K} = f$. In other words, the
- sheaf $\mathcal{C}^{\infty}$ is soft.
-
- In a similar fashion we see that the sheaf of sections of a smooth vector bundle
- on $M$ is soft.
- \end{bsp}
-
- \color{gray}
-
- \begin{bsp}
- If $\mathcal{A}$ is a soft sheaf of rings and $\mathcal{F}$ is a sheaf of $\mathcal{A}$-modules,
- then $\mathcal{F}$ is soft. Indeed, let $s \in \mathcal{F}(K)$ be a section
- over a compact set $K \subseteq X$, i.e. a section on some open neighbourhood of $K$. By
- softness we can extend the section $1 \in \mathcal{A}(K)$ to a compactly supported global section
- $i \in \mathcal{A}(X)$ with support in $U$. Thus
- $si$ extends to a global section of $\mathcal{F}$.
- \end{bsp}
-
- \begin{satz}
- Let $X$ be a space.
- If $\mathcal{F} \in \sh{X}$ is soft, $K \subseteq X$ is compact and $K \subseteq U$ is an open neighbourhood,
- any section over $K$ can be extended to a global section with compact support contained in $U$.
- \end{satz}
-
- \begin{proof}
- Let $s \in \mathcal{F}(K)$.
- By local compactness, there exists a compact neighbourhood $L$ of $K$ with $L \subseteq U$. Then
- $K \cap \partial L = \emptyset$. Consider the section on $K \cup \partial L$ given by
- $s$ on $K$ and zero on $\partial L$. Since $\mathcal{F}$ is soft, this can be extended
- to a global section, and a fortiori to a section $t$ over $L$. Now
- the sections given by $t$ on $L$ and $0$ on $\overline{X \setminus L}$ glue to a compactly
- supported extension of $s$. Since $L \subseteq U$, its support is contained in $U$.
- \end{proof}
-
- \color{black}
-
- \subsection{Compactly supported cohomology}
-
- Let $X$ be a space.
-
- %\begin{definition}
- % Let $U \subseteq X$ be open and $\mathcal{F} \in \sh{X}$. We define
- % $\Gamma_c(U, \mathcal{F})$ as the subgroup of $\Gamma(U, \mathcal{F})$ consisting of
- % sections with compact support.
- %\end{definition}
- %
- %\begin{bem}
- % If $s, t \in \Gamma(U, \mathcal{F})$ have compact support, so does $s + t$. Thus
- % $\Gamma_c(U, \mathcal{F})$ is indeed a subgroup of $\Gamma(U, \mathcal{F})$.
- %
- % Taking $U = X$, this defines a functor $\Gamma_c = \Gamma_c(X, \cdot)\colon \sh{X} \to \mathcal{A}b$
- %\end{bem}
-
- \begin{theorem}[Base change]
- Let $f\colon X \to Y$ be a continuous map of spaces. For
- $\mathcal{F} \in \sh{X}$, there is a natural isomorphism
- \[
- (f_{!}\mathcal{F})_y \simeq \Gamma_c(f^{-1}(y), \mathcal{F}|_{f^{-1}(y)})
- \] for each $y \in Y$.
- \label{thm:base-change}
- \end{theorem}
-
- \begin{proof}
- Denote by $X_y$ the fibre of $f$ over $y$ and by $\mathcal{F}$ the restriction to $X_y$.
- Let $y \in U \subseteq Y$ open. Then consider the natural map
- \begin{salign*}
- (f_{!}\mathcal{F})(U) &\longrightarrow \Gamma_c(X_y, \mathcal{F}_y) \\
- s &\longmapsto s|_{X_y}
- .\end{salign*}
- This is well-defined, since for any $s \in \mathcal{F}(f^{-1}(U))$ with
- $\supp{s} \xrightarrow{f} U$ proper, we have
- \[
- \supp{s|_{X_y}} = \supp{s} \cap X_y = \left( f|_{\supp{s}}^{U} \right)^{-1}(y)
- \] and the right hand side is compact. This map induces
- a natural map
- \[
- (f_{!}\mathcal{F})_y = \colim{y \in U \subseteq Y} (f_{!}\mathcal{F})(U)
- \longrightarrow \Gamma_c(X_y, \mathcal{F}_y)
- .\]
-
- Injectivity: Let $s \in (f_{!}\mathcal{F})(U)$ such that $s|_{X_y} = 0$. Thus
- $s \in \mathcal{F}(f^{-1}(U))$ and $\supp{s} \xrightarrow{f} U$ is proper. Since
- $s|_{X_y} = 0$, $f^{-1}(y) \cap \supp{s} = X_y \cap \supp{s} = \emptyset$, in particular
- $y \not\in f(\supp{s})$. Let $y \in U'$ be the complement of $f(\supp{s})$ in $U$.
- Since $\supp{s} \xrightarrow{f} U$ is proper, $f(\supp{s})$ is closed in $U$, so
- $U'$ is open in $U$ and hence in $Y$. Moreover
- \[
- f^{-1}(U') \cap \supp{s}
- \subseteq f^{-1}(U') \cap f^{-1}(f(\supp{s}))
- = f^{-1}(U' \cap f(\supp{s}))
- = f^{-1}(\emptyset)
- = \emptyset
- .\]
- Hence $s|_{f^{-1}(U')} = 0$, so $s|_{U'} = 0$.
-
- Surjectivity: Suppose first $\mathcal{F}$ is soft and let
- $s \in \Gamma_c(X_y, \mathcal{F}_y)$. Since $\mathcal{F}$ is soft, we may extend
- $s \in \mathcal{F}(X_y)$ to a compactly supported $s \in \mathcal{F}(X) = (f_{*}\mathcal{F})(Y)$.
- Since $Y$ is Hausdorff, every compact $K \subseteq Y$ is closed and therefore its preimage
- under $f|_{\supp{s}}$ is closed in the compact $\supp{s}$, thus itself compact. Hence
- $f|_{\supp{s}}\colon \supp{s} \to Y$ is proper and $s \in (f_{!}\mathcal{F})(Y)$.
-
- For arbitrary $\mathcal{F}$, there exists an exact sequence
- \[
- \begin{tikzcd}
- 0 \arrow{r} & \mathcal{F} \arrow{r}
- & \mathcal{I} \arrow{r}
- & \mathcal{J}
- \end{tikzcd}
- \] with $\mathcal{I}, \mathcal{J}$ soft (e.g. injective). The functors
- $(f_{!} \cdot )_y$ and $\Gamma_c(X_y, \cdot |_{X_y})$ are left exact, so we have a commuting diagram
- with exact rows:
- \[
- \begin{tikzcd}
- 0 \arrow{r} & (f_!\mathcal{F})_y \arrow{r} \arrow{d}
- & (f_!\mathcal{I})_y \arrow{r} \arrow{d}{\simeq}
- & (f_!\mathcal{J})_y \arrow{d}{\simeq} \\
- 0 \arrow{r} & \Gamma_c(X_y, \mathcal{F}_y) \arrow{r}
- & \Gamma_c(X_y, \mathcal{I}_y) \arrow{r}
- & \Gamma_c(X_y, \mathcal{J}_y)
- \end{tikzcd}
- .\] The five-lemma yields the desired isomorphism.
- \end{proof}
-
- \begin{satz}[Lower shriek is exact on soft]
- Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ be an exact sequence
- in $\sh{X}$ with $\mathcal{F}'$ soft. Then the sequence
- \[
- 0 \to f_{!}\mathcal{F}' \to f_{!}\mathcal{F} \to f_{!}\mathcal{F}'' \to 0
- \] is exact.
- \label{satz:lower-shriek-exact-on-soft}
- \end{satz}
-
- \begin{proof}
- Since $f_{!}$ is left exact, we only need to show the surjectivity on the right, i.e.
- for every $y \in Y$ the surjectivity of $(f_{!}\mathcal{F})_y \to (f_{!}\mathcal{F}'')_y$.
- We have the following commutative diagram:
- \[
- \begin{tikzcd}
- \Gamma_c(f^{-1}(y), \mathcal{F}|_{f^{-1}(y)}) \arrow{r} \arrow{d} & \arrow{d}
- \Gamma_c(f^{-1}(y), \mathcal{F}''|_{f^{-1}(y)}) \\
- (f_!\mathcal{F})_y \arrow{r} & (f_!\mathcal{F}'')_y
- \end{tikzcd}
- .\] By \ref{thm:base-change}, the vertical arrows are isomorphisms. It suffices
- thus to show the surjectivity of
- $\Gamma_c(f^{-1}(y), \mathcal{F}_{f^{-1}(y)}) \to \Gamma_c(f^{-1}(y), \mathcal{F}''|_{f^{-1}(y)})$.
- Restriction to $f^{-1}(y)$ is exact, moreover it preserves softness. We thus reduced
- to showing that $\Gamma_c(X, \cdot)$ preserves surjections.
-
- Suppose first that $X$ is compact and let $s \in \Gamma_c(X, \mathcal{F}'') = \Gamma(X, \mathcal{F}'')$.
- Since $\mathcal{F} \to \mathcal{F}'' \to 0$ is exact, there exist
- a covering $X = \bigcup_{i \in I} U_i$ and lifts $t_i \in \mathcal{F}(U_i)$
- of $s|_{U_i}$. By local compactness of $X$, we may assume, after a possible refinement, that each
- $U_i$ contains a compact set $V_i$ whose interiors still cover $X$. Since
- $X$ is compact, we may assume $I$ is finite. To piece together the $t_i$, we may assume, by induction,
- that $\#I = 2$.
-
- Consider $t_1|_{U_1 \cap U_2} - t_2|_{U_1 \cap U_2}$. This is necessarily a section $e'$ of
- $\mathcal{F}'(U_1 \cap U_2)$ as it maps to zero in $\mathcal{F}''(U_1 \cap U_2)$. Restricting
- $e'$ to the compact $V_1 \cap V_2$ and extending it by softness, yields a global section $e$ of
- $\mathcal{F}'$. Now
- \[
- (t_2|_{V_2} + e|_{V_2})|_{V_1 \cap V_2} = t_2|_{V_1 \cap V_2} + e'|_{V_1 \cap V_2} = t_1|_{V_1 \cap V_2}
- .\] Thus $t_1|_{V_1}, t_2|_{V_2} + e|_{V_2}$ glue to a global section $t$ of $\mathcal{F}$
- with image $s$.
-
- Now for general $X$: Let $s \in \mathcal{F}''(X)$ with compact support $Z$. By local compactness,
- there exists a compact neighbourhood $Z' \subseteq X$ of $Z$. Since
- pullback of sheaves is exact and restriction of soft sheaves to closed subsets preserves softness,
- applying the result to $Z'$,
- yields a section $t' \in \mathcal{F}(Z')$ lifting $s|_{Z'}$. The restriction
- $t'|_{\partial Z'}$ maps to $s|_{\partial Z'} = 0$, so $t'|_{\partial Z'} \in \mathcal{F}'(\partial Z')$.
- Since $\partial Z'$ is compact and $\mathcal{F}'$ is soft, $t'|_{\partial Z'}$
- extends to a global section $b$ of $\mathcal{F}'$. Thus
- \[
- (t' - b|_{Z'})|_{\partial Z'} = t'|_{\partial Z'} - t'|_{\partial Z'} = 0
- .\] So
- $t' - b|_{Z'}$ on $Z'$ and $0$ on $\overline{X \setminus Z'}$ glue to a global section
- $t$ of $\mathcal{F}$. Then $t|_{Z'} = t' - b|_{Z'}$ maps to $s|_{Z'}$ since
- $b \in \mathcal{F}'(X)$. Since $\supp{t}, \supp{s} \subseteq Z'$, $t$ is a compactly supported lift of $s$.
- \end{proof}
-
- \begin{korollar}
- If $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ is an exact sequence
- in $\sh{X}$ and $\mathcal{F}', \mathcal{F}$ are soft, then
- $\mathcal{F}''$ is soft too.
- \label{kor:soft-2+3}
- \end{korollar}
-
- \begin{proof}
- Let $Z \subseteq X$ be closed.
- Since restricting to a closed subset is exact and preserves softness,
- by \ref{satz:lower-shriek-exact-on-soft}
- $\Gamma_c(Z, \mathcal{F}) \to \Gamma_c(Z, \mathcal{F}'')$ is surjective.
- This yields a commutative
- diagram
- \[
- \begin{tikzcd}
- \Gamma_c(X, \mathcal{F}) \arrow[twoheadrightarrow]{d} \arrow{r} & \Gamma_c(X, \mathcal{F}'')
- \arrow{d} \\
- \Gamma_c(Z, \mathcal{F}) \arrow[twoheadrightarrow]{r} & \Gamma_c(Z, \mathcal{F}'')
- \end{tikzcd}
- ,\] where the left vertical arrow is surjective, since $\mathcal{F}$ is soft. Since
- the composition is surjective, $\Gamma_c(X, \mathcal{F}'') \to \Gamma_c(Z, \mathcal{F}'')$ is also
- surjective.
- \end{proof}
-
- \section{Derived categories and functors}
-
- We give a brief introduction to the derived category of an abelian category $\mathcal{A}$. Let
- $F\colon \mathcal{A} \to \mathcal{B}$ be a left exact functor and let $\mathcal{A}$ have enough
- injectives. Then the classical derived functors exist. To compute $\mathrm{R}^{i}F(X)$ for an
- object $X \in \mathcal{A}$, we choose an injective resolution
- \[
- \begin{tikzcd}
- 0 \arrow{r} & X \arrow{r} \arrow{d} & 0 \arrow{r} \arrow{d} & 0 \arrow{r} \arrow{d} & \cdots\\
- 0 \arrow{r} & I_0 \arrow{r} & I_1 \arrow{r} & I_2 \arrow{r} & \cdots
- \end{tikzcd}
- \] i.e. a quasiisomorphism $X \to \com{I} $. Then
- $\mathrm{R}^{i}F(X) = H^{i} F(\com{I})$.
-
- New idea: identify $X$ with its resolution, in other words, turn quasiisomorphisms into
- isomorphisms. First step in this direction: Consider the category $\mathcal{K}(\mathcal{A})$
- of complexes where arrows are homomorphisms of complexes up to homotopy. Still
- quasiisomorphisms are in general not isomorphisms, so need to do more:
-
- Localise by the class of quasiisomorphisms. This is then called the derived category
- of $\mathcal{A}$:
- \[
- \mathcal{D}(\mathcal{A}) = \mathcal{K}(\mathcal{A})_{\mathcal{Q}is}
- .\]
- Exactly like in the situation for rings, not every functor $\mathcal{K}(A) \to \mathcal{K}(B)$
- descends to the derived category, since it needs to send quasiisomorphisms to quasiisomorphisms.
- If the functor is induced by an exact functor $\mathcal{A} \to \mathcal{B}$, this is the case. For
- an arbitrary $F\colon \mathcal{A} \to \mathcal{B}$, we can hope that a derived functor exists. This
- is defined by a universal property, that ensures that this derived functor is in a sense
- close to the original one.
-
- For a left exact functor $F\colon \mathcal{A} \to \mathcal{B}$, there is the following result:
-
- \begin{theorem}
- If there exists a full additive subcategory $\mathcal{L}$ in $\mathcal{A}$ that is \emph{adapted} to $F$, i.e.
- \begin{enumerate}[(i)]
- \item for any $X \in \mathcal{A}$ there exists
- $X' \in \mathcal{L}$ and an exact sequence
- $0 \to X \to X'$
- \item if $0 \to X' \to X \to X'' \to 0$ is exact sequence in $\mathcal{A}$ and
- $X'$, $X$ are in $\mathcal{L}$, then $X''$ is in $\mathcal{L}$
- \item if $0 \to X' \to X \to X'' \to 0$ is exact sequence in $\mathcal{A}$ and
- if $X', X, X''$ are in $\mathcal{L}$, then the sequence
- $0 \to F(X') \to F(X) \to F(X'') \to 0$ is exact.
- \end{enumerate}
- Then the derived functor
- $\mathrm{R}F\colon \mathcal{D}^{+}(\mathcal{A}) \to \mathcal{D}^{+}(\mathcal{B})$
- exists and for any $\com{I} \in \mathcal{K}^{+}(\mathcal{L})$ we have a
- natural isomorphism
- \[
- \mathrm{R} F (\com{I}) \simeq F(\com{I})
- .\]
- \end{theorem}
-
- Since $\sh{X}$ has enough injectives and every injective sheaf is soft, by
- \ref{satz:lower-shriek-exact-on-soft} and \ref{kor:soft-2+3}, the
- class of soft sheaves is adapted to the functor $f_!$. Thus the derived functor
-
- \[
- \mathrm{R} f_{!} \colon \mathcal{D}^{+}(X) \longrightarrow \mathcal{D}^{+}(Y)
- \] exists.
-
- \begin{korollar}
- For $\com{\mathcal{F}} \in \mathcal{K}om^{+}(\sh{X})$, we have a natural isomorphism
- \[
- (\mathrm{R}f_{!} \com{\mathcal{F}})_y
- \simeq \mathrm{R} \Gamma_c(f^{-1}(y), \com{\mathcal{F}}|_{f^{-1}(y)})
- \] in $\mathcal{D}(X)$.
- \end{korollar}
-
- \begin{proof}
- Let $\com{\mathcal{F}} \to \com{\mathcal{I}}$ be an injective resolution. Then
- \begin{salign*}
- (\mathrm{R}f_{!} \com{\mathcal{F}})_y
- &\simeq (\mathrm{R}f_{!} \com{\mathcal{I}})_y \\
- &\simeq (f_{!} \com{\mathcal{I}})_y \\
- &\simeq \Gamma_c(f^{-1}(y), \com{\mathcal{I}}|_{f^{-1}(y)}) \\
- &\simeq \mathrm{R}\Gamma_c(f^{-1}(y), \com{\mathcal{I}}|_{f^{-1}(y)}) \\
- &\simeq \mathrm{R}\Gamma_c(f^{-1}(y), \com{\mathcal{F}}|_{f^{-1}(y)})
- .\end{salign*}
- \end{proof}
-
- %\begin{korollar}
- % Soft sheaves are $\Gamma_c$-acyclic.
- % \label{kor:soft-gamma_c-acyclic}
- %\end{korollar}
- %
- %\begin{proof}
- % Let $\mathcal{F} \in \sh{X}$ be soft and
- % embed $\mathcal{F}$ in an injective sheaf $\mathcal{I}$. This yields an exact sequence
- % \[
- % \begin{tikzcd}
- % 0 \arrow{r} & \mathcal{F} \arrow{r}
- % & \mathcal{I} \arrow{r}
- % & \mathcal{G} \arrow{r}
- % & 0
- % \end{tikzcd}
- % .\]
- % Since $\mathcal{I}$ is injective, in particular flasque, hence soft,
- % by \ref{kor:soft-2+3}, $\mathcal{G}$ is soft.
- % We proceed by induction. For $i = 1$ consider the exact sequence
- % \[
- % \begin{tikzcd}
- % 0 \arrow{r} & \Gamma_c(X, \mathcal{F}) \arrow{r}
- % & \Gamma_c(X, \mathcal{I}) \arrow{r}
- % & \Gamma_c(X, \mathcal{G}) \arrow{r}
- % & H_c^{1}(X, \mathcal{F}) \arrow{r}
- % & \underbrace{H_c^{1}(X, \mathcal{I})}_{= 0}
- % \end{tikzcd}
- % .\]
- % Since $\mathcal{F}$ is soft, $\Gamma_c(X, \mathcal{I}) \to \Gamma_c(X, \mathcal{G})$ is
- % surjective. By the exactness of the sequence, $H_c^{1}(X, \mathcal{F})$ vanishes.
- % Now assume $H_c^{i}(X, \mathcal{F}) = 0$ for any soft sheaf $\mathcal{F}$. Then the exact sequence
- % \[
- % \begin{tikzcd}
- % \underbrace{H_c^{i}(X, \mathcal{I})}_{= 0} \arrow{r} & H_c^{i}(X, \mathcal{G}) \arrow{r}
- % & H_c^{i+1}(X, \mathcal{F}) \arrow{r}
- % & \underbrace{H_c^{i+1}(X, \mathcal{I})}_{= 0}
- % \end{tikzcd}
- % \] yields an isomorphism $H_c^{i}(X, \mathcal{G}) \simeq H_c^{i+1}(X, \mathcal{F})$ and
- % since $\mathcal{G}$ is soft, the left hand side is zero by induction hypothesis.
- %\end{proof}
-
- %\begin{satz}
- % Soft sheaves are $f_!$-acyclic. In particular, if
- % $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ is an exact sequence in $\sh{X}$
- % and $\mathcal{F}'$ is soft, then the sequence
- % $0 \to f_!\mathcal{F}' \to f_!\mathcal{F} \to f_!\mathcal{F}'' \to 0$ is exact.
- %\end{satz}
- %
- %\begin{proof}
- % Let $i > 0$ and $\mathcal{F} \in \sh{X}$ be soft. Then for $y \in Y$
- % \begin{salign*}
- % (R^{i}f_!\mathcal{F})_y
- % \stackrel{\ref{thm:base-change}}{\simeq} H_c^{i}(f^{-1}(y), \mathcal{F}|_{f^{-1}(y)})
- % \; \stackrel{\ref{kor:soft-gamma_c-acyclic}}{=} \; 0
- % ,\end{salign*}
- % since the restriction of a soft sheaf to a closed subset is soft.
- %\end{proof}
-
- \color{gray}
-
- \begin{bsp}
- Let $U \subseteq X$ be open and $j\colon U \to X$ the inclusion map. By looking at stalks,
- one finds that $j_!\mathcal{F}$ for $\mathcal{F} \in \sh{U}$ is just extension by zero.
- \end{bsp}
-
- \begin{satz}[Lower shriek preserves softness]
- If $f\colon X \to Y$ is continuous and $\mathcal{F} \in \sh{X}$ is soft, then
- $f_! \mathcal{F}$ is soft too.
- \end{satz}
-
- \begin{proof}
- Let $Z \subseteq Y$ be compact and
- $s \in (f_!\mathcal{F})(Z) \simeq \colim{Z \subseteq U \subseteq Y} (f_!\mathcal{F})(U)$. Then
- there exists an open neighbourhood $U$ of $Z$ and an extension
- $\tilde{s} \in (f_!\mathcal{F})(U) \subseteq \mathcal{F}(f^{-1}(U))$ with
- $\supp{\tilde{s}} \xrightarrow{f} U$ proper. Since $Y$ is locally compact, there exists
- a compact neighbourhood $L \subseteq U$ of $Z$. Restricting $\tilde{s}$ to the compact
- $K \coloneqq \left(f|_{\supp{\tilde{s}}}\right)^{-1}(L) \subseteq \supp{\tilde{s}}$
- and extending by softness of $\mathcal{F}$, yields a compactly supported global section
- $t \in \mathcal{F}(X) = (f_{*}\mathcal{F})(Y)$ such that $t|_Z = s$. Since
- $\supp{t}$ is compact and $Y$ is Hausdorff, $\supp{t} \xrightarrow{f} Y$ is proper.
- \end{proof}
-
- \begin{korollar}[Leray spectral sequence]
- Given continuous maps $f\colon X \to Y$, $g\colon Y \to Z$ of spaces,
- there is a natural isomorphism
- $\mathrm{R}(g \circ f)_{!} \simeq \mathrm{R}g_{!} \circ \mathrm{R}f_{!}$.
- \end{korollar}
-
- \begin{proof}
- Since soft sheaves are $f_{!}$ (and $g_!$) acyclic and $f_{!}$ maps
- soft sheaves to soft sheaves, the result follows from
- Proposition 5.4 in \cite{hartshorne}.
- \end{proof}
-
- \color{black}
-
- \section{Other functors on abelian sheaves}
-
- \begin{tabular}{l|l|l|l}
- Functor & Exactness & Derivative & Adapted class \\ \hline
- $f^* \colon \sh{Y} \to \sh{X}$
- & exact
- & $f^{*}\colon \mathcal{D}(Y) \to \mathcal{D}(X)$ \\
- $f_* \colon \sh{X} \to \sh{Y}$
- & left exact
- & $\mathrm{R} f_{*}\colon \mathcal{D}(X) \to \mathcal{D}(Y)$
- & K-limp complexes \\
- $\cdot \otimes \mathcal{F} \colon \sh{X} \to \sh{X}$
- & right exact
- & $\cdot \otimes^{L} \com{\mathcal{F}} \colon \mathcal{D}(X) \to \mathcal{D}(X)$
- & K-flat complexes \\
- $\underline{\operatorname{Hom}}(\mathcal{F}, \cdot)\colon \sh{X} \to \sh{X}$
- & left exact
- & $\mathrm{R}\com{\underline{\operatorname{Hom}}}(\com{\mathcal{F}}, \cdot)\colon \mathcal{D}(X) \to \mathcal{D}(X)$
- & K-injective complexes \\
- $f_!\colon \sh{X} \to \sh{Y}$
- & left exact
- & $\mathrm{R}f_{!}\colon \mathcal{D}^{+}(X) \to \mathcal{D}^{+}(Y)$
- & soft sheaves \\
- &
- & $f^{!}\colon \mathcal{D}^{+}(Y) \to \mathcal{D}(X)$
- \end{tabular}
-
- \noindent The internal $\mathrm{Hom}$ functor is for $\mathcal{F}, \mathcal{G} \in \sh{X}$
- given by the formula
- \[
- \underline{\operatorname{Hom}}(\mathcal{F}, \mathcal{G})(U)
- = \operatorname{Hom}_{\sh{U}}(\mathcal{F}|_U, \mathcal{G}|_U)
- \] for every $U \subseteq X$ open and the (internal) tensor product by the sheafification
- of the presheaf
- \[
- U \mapsto \mathcal{F}(U) \otimes \mathcal{G}(U)
- .\]
- These functors satisfy the following adjunction results
-
- \[
- f^{*} \dashv \mathrm{R}f_{*}
- \] and
- \[
- \cdot \otimes^{L} \com{\mathcal{F}} \dashv
- \mathrm{R}\underline{\operatorname{Hom}}(\com{\mathcal{F}}, \cdot )
- .\]
-
- \bibliographystyle{alpha}
- \bibliography{refs}
-
- \end{document}
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