# The Stacks Project

## Tag 008C

### 6.21. Continuous maps and sheaves

Let $f : X \to Y$ be a continuous map of topological spaces. We will define the pushforward and pullback functors for presheaves and sheaves.

Let $\mathcal{F}$ be a presheaf of sets on $X$. We define the pushforward of $\mathcal{F}$ by the rule $$f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))$$ for any open $V \subset Y$. Given $V_1 \subset V_2 \subset Y$ open the restriction map is given by the commutativity of the diagram $$\xymatrix{ f_*\mathcal{F}(V_2) \ar[d] \ar@{=}[r] & \mathcal{F}(f^{-1}(V_2)) \ar[d]^{\text{restriction for }\mathcal{F}} \\ f_*\mathcal{F}(V_1) \ar@{=}[r] & \mathcal{F}(f^{-1}(V_1)) }$$ It is clear that this defines a presheaf of sets. The construction is clearly functorial in the presheaf $\mathcal{F}$ and hence we obtain a functor $$f_* : \textit{PSh}(X) \longrightarrow \textit{PSh}(Y).$$

Lemma 6.21.1. Let $f : X \to Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of sets on $X$. Then $f_*\mathcal{F}$ is a sheaf on $Y$.

Proof. This immediately follows from the fact that if $V = \bigcup V_j$ is an open covering in $Y$, then $f^{-1}(V) = \bigcup f^{-1}(V_j)$ is an open covering in $X$. $\square$

As a consequence we obtain a functor $$f_* : \mathop{\textit{Sh}}\nolimits(X) \longrightarrow \mathop{\textit{Sh}}\nolimits(Y).$$ This is compatible with composition in the following strong sense.

Lemma 6.21.2. Let $f : X \to Y$ and $g : Y \to Z$ be continuous maps of topological spaces. The functors $(g \circ f)_*$ and $g_* \circ f_*$ are equal (on both presheaves and sheaves of sets).

Proof. This is because $(g \circ f)_*\mathcal{F}(W) = \mathcal{F}((g \circ f)^{-1}W)$ and $(g_* \circ f_*)\mathcal{F}(W) = \mathcal{F}(f^{-1} g^{-1} W)$ and $(g \circ f)^{-1}W = f^{-1} g^{-1} W$. $\square$

Let $\mathcal{G}$ be a presheaf of sets on $Y$. The pullback presheaf $f_p\mathcal{G}$ of a given presheaf $\mathcal{G}$ is defined as the left adjoint of the pushforward $f_*$ on presheaves. In other words it should be a presheaf $f_p \mathcal{G}$ on $X$ such that $$\mathop{\rm Mor}\nolimits_{\textit{PSh}(X)}(f_p\mathcal{G}, \mathcal{F}) = \mathop{\rm Mor}\nolimits_{\textit{PSh}(Y)}(\mathcal{G}, f_*\mathcal{F}).$$ By the Yoneda lemma this determines the pullback uniquely. It turns out that it actually exists.

Lemma 6.21.3. Let $f : X \to Y$ be a continuous map. There exists a functor $f_p : \textit{PSh}(Y) \to \textit{PSh}(X)$ which is left adjoint to $f_*$. For a presheaf $\mathcal{G}$ it is determined by the rule $$f_p\mathcal{G}(U) = \mathop{\rm colim}\nolimits_{f(U) \subset V} \mathcal{G}(V)$$ where the colimit is over the collection of open neighbourhoods $V$ of $f(U)$ in $Y$. The colimits are over directed partially ordered sets. (The restriction mappings of $f_p\mathcal{G}$ are explained in the proof.)

Proof. The colimit is over the partially ordered set consisting of open subsets $V \subset Y$ which contain $f(U)$ with ordering by reverse inclusion. This is a directed partially ordered set, since if $V, V'$ are in it then so is $V \cap V'$. Furthermore, if $U_1 \subset U_2$, then every open neighbourhood of $f(U_2)$ is an open neighbourhood of $f(U_1)$. Hence the system defining $f_p\mathcal{G}(U_2)$ is a subsystem of the one defining $f_p\mathcal{G}(U_1)$ and we obtain a restriction map (for example by applying the generalities in Categories, Lemma 4.14.7).

Note that the construction of the colimit is clearly functorial in $\mathcal{G}$, and similarly for the restriction mappings. Hence we have defined $f_p$ as a functor.

A small useful remark is that there exists a canonical map $\mathcal{G}(U) \to f_p\mathcal{G}(f^{-1}(U))$, because the system of open neighbourhoods of $f(f^{-1}(U))$ contains the element $U$. This is compatible with restriction mappings. In other words, there is a canonical map $i_\mathcal{G} : \mathcal{G} \to f_* f_p \mathcal{G}$.

Let $\mathcal{F}$ be a presheaf of sets on $X$. Suppose that $\psi : f_p\mathcal{G} \to \mathcal{F}$ is a map of presheaves of sets. The corresponding map $\mathcal{G} \to f_*\mathcal{F}$ is the map $f_*\psi \circ i_\mathcal{G} : \mathcal{G} \to f_* f_p \mathcal{G} \to f_* \mathcal{F}$.

Another small useful remark is that there exists a canonical map $c_\mathcal{F} : f_p f_* \mathcal{F} \to \mathcal{F}$. Namely, let $U \subset X$ open. For every open neighbourhood $V \supset f(U)$ in $Y$ there exists a map $f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))\to \mathcal{F}(U)$, namely the restriction map on $\mathcal{F}$. And this is compatible with the restriction mappings between values of $\mathcal{F}$ on $f^{-1}$ of varying opens containing $f(U)$. Thus we obtain a canonical map $f_p f_* \mathcal{F}(U) \to \mathcal{F}(U)$. Another trivial verification shows that these maps are compatible with restriction maps and define a map $c_\mathcal{F}$ of presheaves of sets.

Suppose that $\varphi : \mathcal{G} \to f_*\mathcal{F}$ is a map of presheaves of sets. Consider $f_p\varphi : f_p \mathcal{G} \to f_p f_* \mathcal{F}$. Postcomposing with $c_\mathcal{F}$ gives the desired map $c_\mathcal{F} \circ f_p\varphi : f_p\mathcal{G} \to \mathcal{F}$. We omit the verification that this construction is inverse to the construction in the other direction given above. $\square$

Lemma 6.21.4. Let $f : X \to Y$ be a continuous map. Let $x \in X$. Let $\mathcal{G}$ be a presheaf of sets on $Y$. There is a canonical bijection of stalks $(f_p\mathcal{G})_x = \mathcal{G}_{f(x)}$.

Proof. This you can see as follows \begin{eqnarray*} (f_p\mathcal{G})_x & = & \mathop{\rm colim}\nolimits_{x \in U} f_p\mathcal{G}(U) \\ & = & \mathop{\rm colim}\nolimits_{x \in U} \mathop{\rm colim}\nolimits_{f(U) \subset V} \mathcal{G}(V) \\ & = & \mathop{\rm colim}\nolimits_{f(x) \in V} \mathcal{G}(V) \\ & = & \mathcal{G}_{f(x)} \end{eqnarray*} Here we have used Categories, Lemma 4.14.9, and the fact that any $V$ open in $Y$ containing $f(x)$ occurs in the third description above. Details omitted. $\square$

Let $\mathcal{G}$ be a sheaf of sets on $Y$. The pullback sheaf $f^{-1}\mathcal{G}$ is defined by the formula $$f^{-1}\mathcal{G} = (f_p\mathcal{G})^\# .$$ Sheafification is a left adjoint to the inclusion of sheaves in presheaves, and $f_p$ is a left adjoint to $f_*$ on presheaves. As a formal consequence we obtain that $f^{-1}$ is a left adjoint of pushforward on sheaves. In other words, $$\mathop{\rm Mor}\nolimits_{\mathop{\textit{Sh}}\nolimits(X)}(f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\rm Mor}\nolimits_{\mathop{\textit{Sh}}\nolimits(Y)}(\mathcal{G}, f_*\mathcal{F}).$$ The formal argument is given in the setting of abelian sheaves in the next section.

Lemma 6.21.5. Let $x \in X$. Let $\mathcal{G}$ be a sheaf of sets on $Y$. There is a canonical bijection of stalks $(f^{-1}\mathcal{G})_x = \mathcal{G}_{f(x)}$.

Proof. This is a combination of Lemmas 6.17.2 and 6.21.4. $\square$

Lemma 6.21.6. Let $f : X \to Y$ and $g : Y \to Z$ be continuous maps of topological spaces. The functors $(g \circ f)^{-1}$ and $f^{-1} \circ g^{-1}$ are canonically isomorphic. Similarly $(g \circ f)_p \cong f_p \circ g_p$ on presheaves.

Proof. To see this use that adjoint functors are unique up to unique isomorphism, and Lemma 6.21.2. $\square$

Definition 6.21.7. Let $f : X \to Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of sets on $X$ and let $\mathcal{G}$ be a sheaf of sets on $Y$. An $f$-map $\xi : \mathcal{G} \to \mathcal{F}$ is a collection of maps $\xi_V : \mathcal{G}(V) \to \mathcal{F}(f^{-1}(V))$ indexed by open subsets $V \subset Y$ such that $$\xymatrix{ \mathcal{G}(V) \ar[r]_{\xi_V} \ar[d]_{\text{restriction of }\mathcal{G}} & \mathcal{F}(f^{-1}V) \ar[d]^{\text{restriction of }\mathcal{F}} \\ \mathcal{G}(V') \ar[r]^{\xi_{V'}} & \mathcal{F}(f^{-1}V') }$$ commutes for all $V' \subset V \subset Y$ open.

Lemma 6.21.8. Let $f : X \to Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of sets on $X$ and let $\mathcal{G}$ be a sheaf of sets on $Y$. There are canonical bijections between the following three sets:

1. The set of maps $\mathcal{G} \to f_*\mathcal{F}$.
2. The set of maps $f^{-1}\mathcal{G} \to \mathcal{F}$.
3. The set of $f$-maps $\xi : \mathcal{G} \to \mathcal{F}$.

Proof. We leave the easy verification to the reader. $\square$

It is sometimes convenient to think about $f$-maps instead of maps between sheaves either on $X$ or on $Y$. We define composition of $f$-maps as follows.

Definition 6.21.9. Suppose that $f : X \to Y$ and $g : Y \to Z$ are continuous maps of topological spaces. Suppose that $\mathcal{F}$ is a sheaf on $X$, $\mathcal{G}$ is a sheaf on $Y$, and $\mathcal{H}$ is a sheaf on $Z$. Let $\varphi : \mathcal{G} \to \mathcal{F}$ be an $f$-map. Let $\psi : \mathcal{H} \to \mathcal{G}$ be an $g$-map. The composition of $\varphi$ and $\psi$ is the $(g \circ f)$-map $\varphi \circ \psi$ defined by the commutativity of the diagrams $$\xymatrix{ \mathcal{H}(W) \ar[rr]_{(\varphi \circ \psi)_W} \ar[rd]_{\psi_W} & & \mathcal{F}(f^{-1}g^{-1}W) \\ & \mathcal{G}(g^{-1}W) \ar[ru]_{\varphi_{g^{-1}W}} }$$

We leave it to the reader to verify that this works. Another way to think about this is to think of $\varphi \circ \psi$ as the composition $$\mathcal{H} \xrightarrow{\psi} g_*\mathcal{G} \xrightarrow{g_*\varphi} g_* f_* \mathcal{F} = (g \circ f)_* \mathcal{F}$$ Now, doesn't it seem that thinking about $f$-maps is somehow easier?

Finally, given a continuous map $f : X \to Y$, and an $f$-map $\varphi : \mathcal{G} \to \mathcal{F}$ there is a natural map on stalks $$\varphi_x : \mathcal{G}_{f(x)} \longrightarrow \mathcal{F}_x$$ for all $x \in X$. The image of a representative $(V, s)$ of an element in $\mathcal{G}_{f(x)}$ is mapped to the element in $\mathcal{F}_x$ with representative $(f^{-1}V, \varphi_V(s))$. We leave it to the reader to see that this is well defined. Another way to state it is that it is the unique map such that all diagrams $$\xymatrix{ \mathcal{F}(f^{-1}V) \ar[r] & \mathcal{F}_x \\ \mathcal{G}(V) \ar[r] \ar[u]^{\varphi_V} & \mathcal{G}_{f(x)} \ar[u]^{\varphi_x} }$$ (for $f(x) \in V \subset Y$ open) commute.

Lemma 6.21.10. Suppose that $f : X \to Y$ and $g : Y \to Z$ are continuous maps of topological spaces. Suppose that $\mathcal{F}$ is a sheaf on $X$, $\mathcal{G}$ is a sheaf on $Y$, and $\mathcal{H}$ is a sheaf on $Z$. Let $\varphi : \mathcal{G} \to \mathcal{F}$ be an $f$-map. Let $\psi : \mathcal{H} \to \mathcal{G}$ be an $g$-map. Let $x \in X$ be a point. The map on stalks $(\varphi \circ \psi)_x : \mathcal{H}_{g(f(x))} \to \mathcal{F}_x$ is the composition $$\mathcal{H}_{g(f(x))} \xrightarrow{\psi_{f(x)}} \mathcal{G}_{f(x)} \xrightarrow{\varphi_x} \mathcal{F}_x$$

Proof. Immediate from Definition 6.21.9 and the definition of the map on stalks above. $\square$

The code snippet corresponding to this tag is a part of the file sheaves.tex and is located in lines 1973–2332 (see updates for more information).

\section{Continuous maps and sheaves}
\label{section-presheaves-functorial}

\noindent
Let $f : X \to Y$ be a continuous map of topological spaces.
We will define the pushforward and pullback functors for
presheaves and sheaves.

\medskip\noindent
Let $\mathcal{F}$ be a presheaf of sets on $X$. We define the
{\it pushforward} of $\mathcal{F}$ by the rule
$$f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))$$
for any open $V \subset Y$.
Given $V_1 \subset V_2 \subset Y$ open the restriction map
is given by the commutativity of the diagram
$$\xymatrix{ f_*\mathcal{F}(V_2) \ar[d] \ar@{=}[r] & \mathcal{F}(f^{-1}(V_2)) \ar[d]^{\text{restriction for }\mathcal{F}} \\ f_*\mathcal{F}(V_1) \ar@{=}[r] & \mathcal{F}(f^{-1}(V_1)) }$$
It is clear that this defines a presheaf of sets. The construction
is clearly functorial in the presheaf $\mathcal{F}$ and hence
we obtain a functor
$$f_* : \textit{PSh}(X) \longrightarrow \textit{PSh}(Y).$$

\begin{lemma}
\label{lemma-pushforward-sheaf}
Let $f : X \to Y$ be a continuous map.
Let $\mathcal{F}$ be a sheaf of sets on $X$.
Then $f_*\mathcal{F}$ is a sheaf on $Y$.
\end{lemma}

\begin{proof}
This immediately follows from the fact that
if $V = \bigcup V_j$ is an open covering in $Y$,
then $f^{-1}(V) = \bigcup f^{-1}(V_j)$ is an open covering in $X$.
\end{proof}

\noindent
As a consequence we obtain a functor
$$f_* : \Sh(X) \longrightarrow \Sh(Y).$$
This is compatible with composition in the following
strong sense.

\begin{lemma}
\label{lemma-pushforward-composition}
Let $f : X \to Y$ and $g : Y \to Z$ be continuous maps
of topological spaces. The functors $(g \circ f)_*$
and $g_* \circ f_*$ are equal (on both presheaves
and sheaves of sets).
\end{lemma}

\begin{proof}
This is because $(g \circ f)_*\mathcal{F}(W) = \mathcal{F}((g \circ f)^{-1}W)$ and
$(g_* \circ f_*)\mathcal{F}(W) = \mathcal{F}(f^{-1} g^{-1} W)$
and $(g \circ f)^{-1}W = f^{-1} g^{-1} W$.
\end{proof}

\noindent
Let $\mathcal{G}$ be a presheaf of sets on $Y$.
The {\it pullback presheaf} $f_p\mathcal{G}$
of a given presheaf $\mathcal{G}$ is defined as the left adjoint
of the pushforward $f_*$ on presheaves. In other words it
should be a presheaf $f_p \mathcal{G}$ on $X$ such that
$$\Mor_{\textit{PSh}(X)}(f_p\mathcal{G}, \mathcal{F}) = \Mor_{\textit{PSh}(Y)}(\mathcal{G}, f_*\mathcal{F}).$$
By the Yoneda lemma this determines the pullback uniquely.
It turns out that it actually exists.

\begin{lemma}
\label{lemma-pullback-presheaves}
Let $f : X \to Y$ be a continuous map.
There exists a functor
$f_p : \textit{PSh}(Y) \to \textit{PSh}(X)$
which is left adjoint to $f_*$. For a presheaf
$\mathcal{G}$ it is determined by the rule
$$f_p\mathcal{G}(U) = \colim_{f(U) \subset V} \mathcal{G}(V)$$
where the colimit is over the collection of open neighbourhoods
$V$ of $f(U)$ in $Y$. The colimits are over
directed partially ordered sets.
(The restriction mappings of $f_p\mathcal{G}$ are explained in the proof.)
\end{lemma}

\begin{proof}
The colimit is over the partially ordered set consisting of open subsets
$V \subset Y$ which contain $f(U)$ with ordering by reverse inclusion.
This is a directed partially ordered set, since if $V, V'$ are in it then
so is $V \cap V'$. Furthermore, if
$U_1 \subset U_2$, then every open neighbourhood of $f(U_2)$
is an open neighbourhood of $f(U_1)$. Hence the system defining
$f_p\mathcal{G}(U_2)$ is a subsystem of the one defining
$f_p\mathcal{G}(U_1)$ and we obtain a restriction map (for
example by applying the generalities in Categories,
Lemma \ref{categories-lemma-functorial-colimit}).

\medskip\noindent
Note that the construction of the colimit is clearly functorial
in $\mathcal{G}$, and similarly for the restriction mappings.
Hence we have defined $f_p$ as a functor.

\medskip\noindent
A small useful remark is that there exists
a canonical map $\mathcal{G}(U) \to f_p\mathcal{G}(f^{-1}(U))$,
because the system of open neighbourhoods
of $f(f^{-1}(U))$ contains the element $U$. This is compatible
with restriction mappings. In other words, there is a
canonical map $i_\mathcal{G} : \mathcal{G} \to f_* f_p \mathcal{G}$.

\medskip\noindent
Let $\mathcal{F}$ be a presheaf of sets on $X$.
Suppose that $\psi : f_p\mathcal{G} \to \mathcal{F}$
is a map of presheaves of sets. The corresponding map
$\mathcal{G} \to f_*\mathcal{F}$ is the map
$f_*\psi \circ i_\mathcal{G} : \mathcal{G} \to f_* f_p \mathcal{G} \to f_* \mathcal{F}$.

\medskip\noindent
Another small useful remark is that there exists a
canonical map $c_\mathcal{F} : f_p f_* \mathcal{F} \to \mathcal{F}$.
Namely, let $U \subset X$ open.
For every open neighbourhood $V \supset f(U)$ in $Y$
there exists a map
$f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))\to \mathcal{F}(U)$,
namely the restriction map on $\mathcal{F}$. And this is compatible
with the restriction mappings between values of $\mathcal{F}$
on $f^{-1}$ of varying opens containing $f(U)$. Thus we obtain
a canonical map $f_p f_* \mathcal{F}(U) \to \mathcal{F}(U)$.
Another trivial verification shows that these maps are compatible
with restriction maps and define a map $c_\mathcal{F}$
of presheaves of sets.

\medskip\noindent
Suppose that $\varphi : \mathcal{G} \to f_*\mathcal{F}$
is a map of presheaves of sets. Consider $f_p\varphi : f_p \mathcal{G} \to f_p f_* \mathcal{F}$.
Postcomposing with $c_\mathcal{F}$ gives the desired map
$c_\mathcal{F} \circ f_p\varphi : f_p\mathcal{G} \to \mathcal{F}$.
We omit the verification that this construction is inverse
to the construction in the other direction given above.
\end{proof}

\begin{lemma}
\label{lemma-stalk-pullback-presheaf}
Let $f : X \to Y$ be a continuous map.
Let $x \in X$. Let $\mathcal{G}$ be a presheaf of sets on $Y$.
There is a canonical bijection of stalks
$(f_p\mathcal{G})_x = \mathcal{G}_{f(x)}$.
\end{lemma}

\begin{proof}
This you can see as follows
\begin{eqnarray*}
(f_p\mathcal{G})_x
& = &
\colim_{x \in U} f_p\mathcal{G}(U) \\
& = &
\colim_{x \in U} \colim_{f(U) \subset V} \mathcal{G}(V) \\
& = &
\colim_{f(x) \in V} \mathcal{G}(V) \\
& = &
\mathcal{G}_{f(x)}
\end{eqnarray*}
Here we have used
Categories, Lemma \ref{categories-lemma-colimits-commute},
and the fact that any $V$ open in $Y$ containing $f(x)$
occurs in the third description above. Details omitted.
\end{proof}

\noindent
Let $\mathcal{G}$ be a sheaf of sets on $Y$.
The {\it pullback sheaf} $f^{-1}\mathcal{G}$ is defined
by the formula
$$f^{-1}\mathcal{G} = (f_p\mathcal{G})^\# .$$
Sheafification is a left adjoint to the inclusion
of sheaves in presheaves, and $f_p$ is a left
adjoint to $f_*$ on presheaves. As a formal consequence
we obtain that $f^{-1}$ is a left adjoint of
pushforward on sheaves. In other words,
$$\Mor_{\Sh(X)}(f^{-1}\mathcal{G}, \mathcal{F}) = \Mor_{\Sh(Y)}(\mathcal{G}, f_*\mathcal{F}).$$
The formal argument is given in the setting of abelian
sheaves in the next section.

\begin{lemma}
\label{lemma-stalk-pullback}
Let $x \in X$. Let $\mathcal{G}$ be a sheaf of sets on $Y$.
There is a canonical bijection of stalks
$(f^{-1}\mathcal{G})_x = \mathcal{G}_{f(x)}$.
\end{lemma}

\begin{proof}
This is a combination of Lemmas \ref{lemma-stalk-sheafification}
and \ref{lemma-stalk-pullback-presheaf}.
\end{proof}

\begin{lemma}
\label{lemma-pullback-composition}
Let $f : X \to Y$ and $g : Y \to Z$ be continuous maps
of topological spaces. The functors $(g \circ f)^{-1}$
and $f^{-1} \circ g^{-1}$ are canonically isomorphic.
Similarly $(g \circ f)_p \cong f_p \circ g_p$ on
presheaves.
\end{lemma}

\begin{proof}
To see this
use that adjoint functors are unique up to unique isomorphism,
and Lemma \ref{lemma-pushforward-composition}.
\end{proof}

\begin{definition}
\label{definition-f-map}
Let $f : X \to Y$ be a continuous map.
Let $\mathcal{F}$ be a sheaf of sets on $X$ and
let $\mathcal{G}$ be a sheaf of sets on $Y$.
An {\it $f$-map $\xi : \mathcal{G} \to \mathcal{F}$}
is a collection of maps
$\xi_V : \mathcal{G}(V) \to \mathcal{F}(f^{-1}(V))$
indexed by open subsets $V \subset Y$ such that
$$\xymatrix{ \mathcal{G}(V) \ar[r]_{\xi_V} \ar[d]_{\text{restriction of }\mathcal{G}} & \mathcal{F}(f^{-1}V) \ar[d]^{\text{restriction of }\mathcal{F}} \\ \mathcal{G}(V') \ar[r]^{\xi_{V'}} & \mathcal{F}(f^{-1}V') }$$
commutes for all $V' \subset V \subset Y$ open.
\end{definition}

\begin{lemma}
\label{lemma-f-map}
Let $f : X \to Y$ be a continuous map.
Let $\mathcal{F}$ be a sheaf of sets on $X$ and
let $\mathcal{G}$ be a sheaf of sets on $Y$.
There are canonical bijections between the following three sets:
\begin{enumerate}
\item The set of maps $\mathcal{G} \to f_*\mathcal{F}$.
\item The set of maps $f^{-1}\mathcal{G} \to \mathcal{F}$.
\item The set of $f$-maps $\xi : \mathcal{G} \to \mathcal{F}$.
\end{enumerate}
\end{lemma}

\begin{proof}
We leave the easy verification to the reader.
\end{proof}

\noindent
It is sometimes convenient to think about $f$-maps
instead of maps between sheaves either on $X$ or on $Y$.
We define composition of $f$-maps as follows.

\begin{definition}
\label{definition-composition-f-maps}
Suppose that $f : X \to Y$ and $g : Y \to Z$ are continuous
maps of topological spaces. Suppose that $\mathcal{F}$ is
a sheaf on $X$, $\mathcal{G}$ is a sheaf on $Y$, and
$\mathcal{H}$ is a sheaf on $Z$.
Let $\varphi : \mathcal{G} \to \mathcal{F}$ be an $f$-map.
Let $\psi : \mathcal{H} \to \mathcal{G}$ be an $g$-map.
The {\it composition of $\varphi$ and $\psi$} is the
$(g \circ f)$-map $\varphi \circ \psi$ defined
by the commutativity of the diagrams
$$\xymatrix{ \mathcal{H}(W) \ar[rr]_{(\varphi \circ \psi)_W} \ar[rd]_{\psi_W} & & \mathcal{F}(f^{-1}g^{-1}W) \\ & \mathcal{G}(g^{-1}W) \ar[ru]_{\varphi_{g^{-1}W}} }$$
\end{definition}

\noindent
We leave it to the reader to verify that this works.
$\varphi \circ \psi$ as the composition
$$\mathcal{H} \xrightarrow{\psi} g_*\mathcal{G} \xrightarrow{g_*\varphi} g_* f_* \mathcal{F} = (g \circ f)_* \mathcal{F}$$
Now, doesn't it seem that thinking about $f$-maps is somehow
easier?

\medskip\noindent
Finally, given a continuous map $f : X \to Y$, and an
$f$-map $\varphi : \mathcal{G} \to \mathcal{F}$ there is
a natural map on stalks
$$\varphi_x : \mathcal{G}_{f(x)} \longrightarrow \mathcal{F}_x$$
for all $x \in X$. The image of a representative $(V, s)$
of an element in $\mathcal{G}_{f(x)}$ is mapped to the
element in $\mathcal{F}_x$ with representative $(f^{-1}V, \varphi_V(s))$. We leave it to the reader to see that this
is well defined. Another way to state it is that it is the
unique map such that all diagrams
$$\xymatrix{ \mathcal{F}(f^{-1}V) \ar[r] & \mathcal{F}_x \\ \mathcal{G}(V) \ar[r] \ar[u]^{\varphi_V} & \mathcal{G}_{f(x)} \ar[u]^{\varphi_x} }$$
(for $f(x) \in V \subset Y$ open) commute.

\begin{lemma}
\label{lemma-compose-f-maps-stalks}
Suppose that $f : X \to Y$ and $g : Y \to Z$ are continuous
maps of topological spaces. Suppose that $\mathcal{F}$ is
a sheaf on $X$, $\mathcal{G}$ is a sheaf on $Y$, and
$\mathcal{H}$ is a sheaf on $Z$.
Let $\varphi : \mathcal{G} \to \mathcal{F}$ be an $f$-map.
Let $\psi : \mathcal{H} \to \mathcal{G}$ be an $g$-map.
Let $x \in X$ be a point. The map on stalks
$(\varphi \circ \psi)_x : \mathcal{H}_{g(f(x))} \to \mathcal{F}_x$ is the composition
$$\mathcal{H}_{g(f(x))} \xrightarrow{\psi_{f(x)}} \mathcal{G}_{f(x)} \xrightarrow{\varphi_x} \mathcal{F}_x$$
\end{lemma}

\begin{proof}
Immediate from Definition \ref{definition-composition-f-maps}
and the definition of the map on stalks above.
\end{proof}

Comment #2528 by Dion on May 1, 2017 a 10:02 pm UTC

Is there a mistake in the comment"(for....)" in the last sentence above Lemma 6.21.10.? I think it's f(x) (not x) which is in the open set V.

Comment #2564 by Johan (site) on May 25, 2017 a 6:35 pm UTC

Thanks Dion. Fixed here.

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