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 $x \in V \subset Y$ 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.
    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?
    
    \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 $x \in V \subset Y$ 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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 008C

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?