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244 changes: 244 additions & 0 deletions sheaves.tex
Original file line number Diff line number Diff line change
Expand Up @@ -3218,6 +3218,8 @@ \section{Morphisms of ringed spaces and modules}
is an $\mathcal{O}_Y(V)$-module map. Here we think of
$\mathcal{F}(f^{-1}V)$ as an $\mathcal{O}_Y(V)$-module
via the map $f^\sharp_V : \mathcal{O}_Y(V) \to \mathcal{O}_X(f^{-1}V)$.
In other words, the formula $\varphi_V(as)=f^\sharp_V(a)\varphi_V(s)$ holds,
for $a\in\mathcal{O}_Y(V)$ and $s\in\mathcal{G}(V)$.
The set of $f$-maps between
$\mathcal{G}$ and $\mathcal{F}$ will be in canonical bijection
with the sets
Expand Down Expand Up @@ -3264,7 +3266,249 @@ \section{Morphisms of ringed spaces and modules}
for example.
\end{proof}

\section{Pulling back sections}
\label{section-pullback-section}

\begin{definition}
\label{definition-inverse-image-section}
Let $f:X\to Y$ be a continuous map of topological spaces.
Let $\mathcal{G}$ be a sheaf of sets over $Y$.
Let $V\subset Y$ be an open subset.
Let $s\in\mathcal{G}(V)$ be a section.
The \emph{inverse image section of $s$ along $f$}, denoted $f^{-1}s$,
is the section in $f^{-1}\mathcal{G}(f^{-1}(V))$ given by the
image of $s$ along the unit $\mathcal{G}\to f_*f^{-1}\mathcal{G}$ of the
adjunction between $f_*$ and $f^{-1}$ (see Section \ref{section-presheaves-functorial}).
\end{definition}

The inverse image section is compatible with the adjunction
between $f_*$ and $f^{-1}$ in the following sense.
Let $\mathcal{F}$, $\mathcal{G}$ be sheaves respectively over $X$ and $Y$,
and let $\varphi^\sharp:\mathcal{G}\to f_*\mathcal{F}$ be a morphism of sheaves
whose adjunct is $\varphi^\flat:f^{-1}\mathcal{G}\to \mathcal{F}$.
Then, for $s\in \mathcal{G}(V)$, we have $\varphi^\sharp_V(s)=\varphi^\flat_{f^{-1}(V)}(f^{-1}s)$.
This follows from the fact that
$$
\varphi^\sharp:\mathcal{G}\to f_*f^{-1}\mathcal{G}\xrightarrow{f_*\varphi^\flat}f_*\mathcal{F},
$$
where the first morphism is the unit of the adjunction between $f^{-1}$ and $f_*$.

\begin{lemma}
\label{lemma-inverse-image-section-germ}
Let $f:X\to Y$ be a continuous map of topological spaces.
Let $\mathcal{G}$ be a sheaf of sets over $Y$.
Let $V\subset Y$ be an open subset.
Let $s\in\mathcal{G}(V)$ be a section.
Let $x\in X$ be a point.
Then $(f^{-1}s)_x=s_{f(x)}$ via the isomorphism of Lemma \ref{lemma-stalk-pullback}.
\end{lemma}

\begin{proof}
Since $\mathcal{G}\to f_*f^{-1}\mathcal{G}$ is an $f$-map,
it follows from the commutative square to be found before Lemma \ref{lemma-compose-f-maps-stalks}.
\end{proof}

Suppose $\mathcal{O}$ is a sheaf of rings over $Y$ and $\mathcal{G}$ is an $\mathcal{O}$-module.
From Lemma \ref{lemma-inverse-image-section-germ} it follows that
$f^{-1}(as)=f^{-1}(a)f^{-1}(s)$
for $a\in\mathcal{O}(V)$ and $s\in\mathcal{G}(V)$,
where we are using the $f^{-1}\mathcal{O}$-module structure on
$f^{-1}\mathcal{G}$ coming from Lemma \ref{lemma-pullback-module}.

\begin{lemma}
\label{lemma-inverse-image-section-morphism}
Let $f :X\to Y$ be a continuous map of topological spaces.
Let $\varphi:\mathcal{G}\to\mathcal{G}'$ be a morphism of
sheaves of sets over $Y$.
Let $V\subset Y$ be an open subset.
Let $s\in\mathcal{G}(V)$ be a section.
Then $f^{-1}(\varphi(s))=(f^{-1}\varphi)(f^{-1}s)$.
\end{lemma}

\begin{proof}
This follows from the naturality of the unit, i.e.,
the following square commutes:
$$
\xymatrix{
\mathcal{G} \ar@{->}[r] \ar@{->}[d] & \mathcal{G}' \ar@{->}[d] \\
f_*f^{-1}\mathcal{G} \ar@{->}[r] & f_*f^{-1}\mathcal{G}'
}
$$
In the square, if we map $s$ first down then right,
we get $(f^{-1}\varphi)(f^{-1}s)$. If we map $s$ first right then down,
we get $f^{-1}(\varphi(s))$.
\end{proof}

\begin{lemma}
\label{lemma-inverse-image-section-functorial}
Let
$X\xrightarrow{f}Y\xrightarrow{g}Z$ be continuous maps of topological spaces.
Let $\mathcal{H}$ be a sheaf of sets over $Z$.
Let $W\subset Z$ be an open subset.
Let $t\in\mathcal{G}(W)$ be a section.
By the means of the isomorphism of functors
$(g\circ f)^{-1}\cong f^{-1}\circ g^{-1}$
of Lemma \ref{lemma-pullback-composition},
we can identify $(g\circ f)^{-1}t=f^{-1}(g^{-1}t)$.
\end{lemma}

\begin{proof}
It suffices to see that $(g\circ f)^{-1}t$ and $f^{-1}(g^{-1}t)$ have same stalk at $x\in X$.
By Lemma \ref{lemma-inverse-image-section-germ},
$((g\circ f)^{-1}t)_x=t_{g(f(x))}=(g^{-1}t)_{f(x)}=f^{-1}(g^{-1}t)_x$.
\end{proof}

\begin{lemma}
\label{lemma-pullback-section}
Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ be a morphism of
ringed spaces. Let $\mathcal{G}$ be a sheaf of
$\mathcal{O}_Y$-modules. Let $V\subset Y$ be an open subset. Let
$s\in\mathcal{G}(V)$ be a section. Denote $g:f^{-1}(V)\to V$ to the
restriction of $f$. The following sections in $f^*\mathcal{G}(f^{-1}(V))$
are the same:
\begin{enumerate}
\item The image of $s$ along the unit
$\mathcal{G}\to f_*f^*\mathcal{G}$ of the adjunction between
$f_*$ and $f^*$ (Lemma \ref{lemma-adjoint-pullback-pushforward-modules}),
\item The global section of $g^*(\mathcal{G}|_V)$ associated with
$\mathcal{O}_{f^{-1}(V)}\cong
g^*\mathcal{O}_V\to g^*(\mathcal{G}|_V)$,
where $\mathcal{O}_V\to\mathcal{G}|_V$ is the map associated with $s$
(see first paragraph of Modules, Section \ref{modules-section-sections}), and
\item The image of $1\otimes f^{-1}s$ (see Definition \ref{definition-inverse-image-section}) along the morphism
$
\mathcal{O}_X\otimes_{p,f^{-1}\mathcal{O}_Y}f^{-1}\mathcal{G}
\to
\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_Y}f^{-1}\mathcal{G}=f^*\mathcal{G}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Naturality of the unit of the adjunction between $g_*$ and $g^*$ gives a
commutative diagram
$$
\xymatrix{
\mathcal{O}_V \ar@{->}[r] \ar@{->}[d] & \mathcal{G}|_V \ar@{->}[d] \\
g_*\mathcal{O}_{f^{-1}(V)}\cong g_*g^*\mathcal{O}_V \ar@{->}[r]
& g_*g^*(\mathcal{G}|_V)
}
$$
The equality between (1) and (2) follows then from the fact that the restriction of the unit
$\mathcal{G}\to f_*f^*\mathcal{G}$ of the adjunction between $f_*$ and $f^*$ to $V$
equals $\mathcal{G}|_V\to g_*g^*(\mathcal{G}|_V)$,
the unit of the adjunction between $g_*$ and $g^*$.

On the other hand, from the proof of Lemma \ref{lemma-adjoint-pullback-pushforward-modules}
it follows that the unit of the adjunction between $f_*$ and $f^*$ at $\mathcal{G}$ is the composite
$$
\mathcal{G}
\xrightarrow{\eta^\mathrm{ip}_\mathcal{G}}
f_*f^{-1}\mathcal{G}
\xrightarrow{f_*(\eta^\mathrm{er}_{f^{-1}\mathcal{G}})}
f_*(\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_Y}f^{-1}\mathcal{G})
$$
where $\eta^\mathrm{ip}$ is the unit of the adjunction between
inverse image and pushforward functors from Lemma \ref{lemma-adjoint-push-pull-modules} and
$\eta^\mathrm{er}$ is the unit of the adjunction between
extension and restriction of scalars functors from Lemma \ref{lemma-adjointness-tensor-restrict}.
Then $s$ maps along this composite as $s\mapsto f^{-1}s\mapsto 1\otimes f^{-1}s$,
which shows the equality between (1) and (3).
\end{proof}

\begin{definition}
\label{definition-pullback-section}
Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ be a morphism of
ringed spaces. Let $\mathcal{G}$ be a sheaf of
$\mathcal{O}_Y$-modules. Let $V\subset Y$ be an open subset. Let
$s\in\mathcal{G}(V)$ be a section. The {\it pullback section of $s$
along $f$}, denoted $f^*s$, is defined to be the section in
$f^*\mathcal{G}(f^{-1}(V))$ from Lemma \ref{lemma-pullback-section}.
\end{definition}

Note that when $\mathcal{G}=\mathcal{O}_Y$,
the unit $\mathcal{O}_Y\to f_*f^*\mathcal{O}_Y\cong f_*\mathcal{O}_X$
is just the map $f^\sharp$ on structure sheaves.
Hence $f^*a=f^\sharp a$ for $a\in\mathcal{O}_Y(V)$.
Note that the unit $\mathcal{G}\to f_*f^*\mathcal{G}$ can be understood as
an $f$-map $\mathcal{G}\to f^*\mathcal{G}$ of sheaves of modules
(see Section \ref{section-ringed-spaces-functoriality-modules});
thus $f^*(as)=f^\sharp(a)f^*(s)$ for $a\in\mathcal{O}_Y(V)$ and $s\in\mathcal{G}(V)$.

The pullback section is compatible with the adjunction
between $f_*$ and $f^*$ in the following sense.
Let $\mathcal{F}$ be an $\mathcal{O}_X$-module, $\mathcal{G}$ be an $\mathcal{O}_Y$-module
and $\varphi^\sharp:\mathcal{G}\to f_*\mathcal{F}$ be a morphism of $\mathcal{O}_Y$-modules
whose adjunct is $\varphi^\flat:f^*\mathcal{G}\to \mathcal{F}$.
Then, for $s\in \mathcal{G}(V)$, we have $\varphi^\sharp_V(s)=\varphi^\flat_{f^{-1}(V)}(f^*s)$.
This follows from the fact that
$$
\varphi^\sharp:\mathcal{G}\to f_*f^*\mathcal{G}\xrightarrow{f_*\varphi^\flat}f_*\mathcal{F},
$$
where the first morphism is the unit of the adjunction between $f^*$ and $f_*$.

\begin{lemma}
\label{lemma-pullback-section-germ}
Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ be a morphism of
ringed spaces. Let $\mathcal{G}$ be a sheaf of
$\mathcal{O}_Y$-modules.
Let $V\subset Y$ be an open subset.
Let $s\in\mathcal{G}(V)$ be a section.
Let $x\in f^{-1}(V)$ be a point.
By means of the isomorphism in
Lemma \ref{lemma-stalk-pullback-modules},
we can identify $(f^*s)_x=s_{f(x)}\otimes 1$.
\end{lemma}

\begin{proof}
Via the isomorphism of Lemma \ref{lemma-stalk-pullback-modules}, we have
$(f^*s)_x=(f^{-1}s\otimes 1)_x=(f^{-1}s)_x\otimes 1=s_{f(x)}\otimes 1$,
by Lemma \ref{lemma-inverse-image-section-germ}.
\end{proof}

\begin{lemma}
\label{lemma-pullback-section-morphism}
Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$
be a morphism of ringed spaces.
Let $\varphi:\mathcal{G}\to\mathcal{G}'$ be a morphism of
sheaves of $\mathcal{O}_Y$-modules.
Let $V\subset Y$ be an open subset.
Let $s\in\mathcal{G}(V)$ be a section.
Then $f^*(\varphi(s))=(f^*\varphi)(f^*s)$.
\end{lemma}

\begin{proof}
This follows from the naturality of the unit, i.e.,
the following square commutes:
$$
\xymatrix{
\mathcal{G} \ar@{->}[r] \ar@{->}[d] & \mathcal{G}' \ar@{->}[d] \\
f_*f^*\mathcal{G} \ar@{->}[r] & f_*f^*\mathcal{G}'
}
$$
In the square, if we map $s$ first down then right,
we get $(f^*\varphi)(f^*s)$. If we map $s$ first right then down,
we get $f^*(\varphi(s))$.
\end{proof}

\begin{lemma}
\label{lemma-pullback-section-functorial}
Let
$(X,\mathcal{O}_X)
\xrightarrow{f}(Y,\mathcal{O}_Y)
\xrightarrow{g}(Z,\mathcal{O}_Z)$ be morphisms of ringed spaces.
Let $\mathcal{H}$ be a sheaf of $\mathcal{O}_Z$-modules.
Let $W\subset Z$ be an open subset.
Let $t\in\mathcal{G}(W)$ be a section.
By the means of the isomorphism of functors
$(g\circ f)^*\cong f^*\circ g^*$
of Lemma \ref{lemma-push-pull-composition-modules},
we can identify $(g\circ f)^*t=f^*(g^*t)$.
\end{lemma}

\begin{proof}
This follows by the description of the pullback section
in point (2) of Lemma \ref{lemma-pullback-section}.
\end{proof}

\section{Skyscraper sheaves and stalks}
\label{section-skyscraper-sheaves}
Expand Down