\documentclass[a4paper]{../../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)}

\begin{document}

\section{Overview}

These notes mostly follow \cite{mathew}. Some ideas are taken
from \cite{gelfand}.

In the following, for a topological space $X$ denote by $\sh{X}$ the category
of sheaves of abelian groups on $X$. Furthermore, denote by
$\mathrm{D}^{+}(X)$ the bounded below derived category of $\sh{X}$.

\begin{definition}[Lower Shriek]
    Let $f\colon X \to Y$ be a continuous map of locally compact topological 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{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}

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 show the existence of the derivative of $f_{!}$, we need to introduce an adapted class of shaves.

\begin{definition}
    Let $X$ be a locally compact space, $\mathcal{F} \in \sh{X}$ and $Z \subseteq X$ a subset. Then
    define
    \[
    \mathcal{F}(Z) = \Gamma(Z, \mathcal{F}) = \Gamma(Z, i^{*}\mathcal{F})
    \] for $i\colon Z \to X$ the canonical inclusion.
\end{definition}

\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 locally compact Hausdorff 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}

\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{satz}
    Let $X$ be a locally compact topological 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}

\subsection{Compactly supported cohomology}

Let $X$ be a topological space.

\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}

\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{bem}[Lower shriek and compact support]
    Let $f\colon X \to \{ *\} $ be the unique continuous map from $X$ to the one point space.
    Then $f_{!} \cdot = \Gamma_c(X, \cdot)$
\end{bem}

\begin{satz}
    $\Gamma_c$ is left exact.
    \label{satz:gamma_c-left-exact}
\end{satz}

\begin{proof}
    Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}''$ be an exact sequence
    in $\sh{X}$. This induces a commutative diagram
    \[
    \begin{tikzcd}
        0 \arrow{r} & \Gamma(X, \mathcal{F}') \arrow{r}
                    & \Gamma(X, \mathcal{F}) \arrow{r}
                    & \Gamma(X, \mathcal{F}'') \\
        0 \arrow{r} & \Gamma_c(X, \mathcal{F}') \arrow{r} \arrow[hookrightarrow]{u}
                    & \Gamma_c(X, \mathcal{F}) \arrow{r} \arrow[hookrightarrow]{u}
                    & \Gamma_c(X, \mathcal{F}'') \arrow[hookrightarrow]{u}
    \end{tikzcd}
    ,\] where the first row is exact. Since the vertical arrows are inclusions,
    the injectivity of $\Gamma_c(X, \mathcal{F}') \to \Gamma_c(X, \mathcal{F})$ is immediate. Let now
    $s \in \Gamma_c(X, \mathcal{F}) \subseteq \Gamma(X, \mathcal{F})$
    such that $s$ becomes zero in $\Gamma_c(X, \mathcal{F}'')$. Thus
    by exactness of the first row, $s \in \Gamma(X, \mathcal{F}')$. Since $s \in \Gamma_c(X, \mathcal{F})$,
    $s$ is compactly supported, so $s \in \Gamma_c(X, \mathcal{F}')$.
\end{proof}

\begin{satz}
    Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ be an exact sequence
    in $\sh{X}$. Suppose $\mathcal{F}'$ is soft. Then the sequence
    $0 \to \Gamma_c(X, \mathcal{F}') \to \Gamma_c(X, \mathcal{F}) \to \Gamma_c(X, \mathcal{F}'') \to 0$
    is also exact.

    \label{satz:soft-gamma_c-exact}
\end{satz}

\begin{proof}
    By \ref{satz:gamma_c-left-exact}, we only need to show surjectivity on the right.

    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 compact. 
    Since restricting to a closed subset is exact and preserves softness,
    by \ref{satz:soft-gamma_c-exact} $\Gamma_c(Z, \mathcal{F}) \to \Gamma_c(Z, \mathcal{F}'')$ 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}

\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{theorem}
    Let $f\colon X \to Y$ be a continuous map of locally compact topological spaces. If $Y$ is Hausdorff and
    $\mathcal{F} \in \sh{X}$, then there is a natural isomorphism
    \[
        (R^{i}f_{!}\mathcal{F})_y \simeq H_c^{i}(f^{-1}(y), \mathcal{F}|_{f^{-1}(y)})
    \] 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 $y \in Y$. Since $R^{i}f_{!}$ is a derived functor, it is a universal $\delta$-functor. Since restriction
    of soft sheaves to closed subspaces preserves softness, the $\delta$-functor
    $\mathcal{F} \mapsto H_c^{i}(X_y, \mathcal{F}_y)$ vanishes for soft sheaves and $i > 0$. Thus
    it is effaceable and hence universal. Therefore it suffices to define a natural isomorphism
    in degree $0$.

    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{theorem}
    Consider a cartesian diagram of locally compact Hausdorff spaces:
    \[
    \begin{tikzcd}
        X \times_Y Z \arrow{r}{f'} \arrow{d}{p'} & X \arrow{d}{p} \\
        Z \arrow{r}{f} & Y
    \end{tikzcd}
    .\] Then there is a natural isomorphism, for any
    $\com{\mathcal{F}} \in \mathcal{D}^{+}(X)$,
    \[
    f^{*} \mathrm{R}p_{!} \com{\mathcal{F}} \simeq \mathrm{R}p_!' f'^{*} \com{\mathcal{F}}
    .\] 
\end{theorem}

\begin{proof}
    By the universal property of derived functors, it suffices to define a natural transformation
    $f^{*}p_{!} \to \mathrm{R} p_{!}'f'^{*}$. By composing with the canonical
    natural transformation $p_{!}'f'^{*} \to \mathrm{R}p_{!}'f'^{*}$, it suffices to define
    the dotted arrow in the diagram below
    \[
    \begin{tikzcd}
        f^{*}p_{!} \arrow[dashed]{rr} \arrow[dotted]{dr} & & \mathrm{R} p_{!}'f'^{*} \\
                                     & p_{!}'f'^{*} \arrow[swap]{ur}{can} &
    \end{tikzcd}
    .\] By naturality, it is sufficient to define for $\mathcal{G} \in \sh{X}$ a natural map
    $f^{*}p_! \mathcal{G} \to p_!'f'^{*}\mathcal{G}$. Since
    $f^{*} \dashv f_{*}$, this is equivalent to defining a natural map
    $p_!\mathcal{G} \to f_{*} p_{!}'f'^{*} \mathcal{G}$.

    Again using $f'^{*} \dashv f'_{*}$, the map $\text{id}_{f'^{*} \mathcal{G}}$ induces a map
    $\mathcal{G} \to f'_{*} f'^{*} \mathcal{G}$. Applying
    $p_{*}$ yields $p_{*} \mathcal{G} \to p_{*}f'_{*}f'^{*} \mathcal{G}$. By the commutativity of the diagram
    we have $p_{*} f'_{*} = (pf')_{*} = (fp')_{*} = f_{*} p'_{*}$, so a map
    $\varphi\colon p_{*} \mathcal{G} \to f_{*} p'_{*} f'^{*} \mathcal{G}$.

    For $U \subseteq Y$ open, this induces a map
    \[
    \varphi_U\colon \mathcal{G}(p^{-1}(U)) \longrightarrow (f'^{*} \mathcal{G})(p'^{-1}(f^{-1}(U)))
    .\]
    Let now $s \in \mathcal{G}(p^{-1}(U))$ such that
    $\supp{s} \xrightarrow{p} U$ is proper. Since $f'^{*}$ preserves stalks, for
    $(x, z) \in p^{-1}(U) \times_U f^{-1}(U)$ we have the following equivalences
    \[
        (x, z) \in \supp{\varphi_U(s)}
        \iff \varphi_U(s)_{(x, z)} \neq 0
        \iff s_{f'(x,z)} \neq 0
        \iff s_{x} \neq 0
        \iff x \in \supp{s}
    .\] Thus $\supp{\varphi_U(s)} = \supp{s} \times_{U} f^{-1}(U)$. We therefore have the following
    commutative diagram:
    \[
    \begin{tikzcd}
        \supp{s} \times_{U} f^{-1}(U) \arrow{d} \arrow{r} & \supp{s} \arrow{d} \\
        f^{-1}(U) \arrow{r} & U
    \end{tikzcd}
    .\] By assumption the right vertical arrow is proper. Since properness is stable under (topological)
    base change, the left vertical arrow is proper too. Hence
    $\supp{\varphi_U(s)} \xrightarrow{p'} f^{-1}(U)$ is proper and
    \[
    \varphi_U(s) \in (p'_{!}f'^{*} \mathcal{G})(f^{-1}(U)) = (f_{*} p'_{!}f'^{*} \mathcal{G})(U)
    .\] Thus $\varphi$ restricts to a natural map
    \[
    p_{!} \mathcal{G} \longrightarrow f_{*} p'_{!} f'^{*} \mathcal{G}
    .\]

    To check that this is an isomorphism, we can use the fact that both functors are
    way-out functors in the sense of Section 7 in \cite{hartshorne}. Thus we only need to check
    this for a single sheaf $\mathcal{F} \in \sh{X}$, i.e. we want to show
    \[
    f^{*} R^{i} p_{!} \mathcal{F} \xlongrightarrow{\simeq} R^{i}p_{!}'f'^{*}\mathcal{F}
    \] for all $i \ge 0$. Again by universality of the $\delta$-functors involved,
    we may assume $i = 0$. Moreover, we can check this at the level of stalks. Let $z \in Z$. Then
    on the left hand side
    \begin{equation}
        (f^{*}p_{!}\mathcal{F})_z
        \simeq
        (p_{!} \mathcal{F})_{f(z)}
        \stackrel{\ref{thm:base-change}}{\simeq}
        \Gamma_c(p^{-1}(f(z)), \mathcal{F}|_{p^{-1}(f(z))})
        =
        \Gamma_c(f'(p'^{-1}(z))), \mathcal{F}|_{f'(p'^{-1}(z))})
        \label{eq:1}
    \end{equation}
    On the right hand side, we have
    \begin{equation}
        (p'_{!} f'^{*} \mathcal{F})_z
        \stackrel{\ref{thm:base-change}}{\simeq}
        \Gamma_c(p'^{-1}(z), (f'^{*} \mathcal{F})|_{p'^{-1}(z)})
        \label{eq:2}
    \end{equation}
    $\mathcal{F}|_{f'(p'^{-1}(z))}$ and
    $(f'^{*} \mathcal{F})|_{p'^{-1}(z)}$ are given as the sheafification of the same presheaf, indeed:
    \begin{salign*}
    \colim{p'^{-1}(z) \subseteq U \subseteq X \times_Y Z} \; (f'^{*}\mathcal{F})(U)
    &= \colim{p'^{-1}(z) \subseteq U \subseteq X \times_Y Z} \quad
        \colim{f'(U) \subseteq V \subseteq X} \; \mathcal{F}(V) \\
    &= \colim{f'(p'^{-1}(z)) \subseteq V \subseteq X} \; \mathcal{F}(V)
    .\end{salign*}
    This shows (\refeq{eq:1}) $\simeq$ (\refeq{eq:2}) and concludes the proof.
\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}

\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 maps $f\colon X \to Y$, $g\colon Y \to Z$ of locally compact Hausdorff 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}

\bibliographystyle{alpha}
\bibliography{refs}

\end{document}