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 $Y$.

## 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

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

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

**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

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

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

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 4.14.8).

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

Here we have used Categories, Lemma 4.14.10, 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

The pullback $f^{-1}$ is a left adjoint of pushforward on sheaves. In other words,

Namely, we have

For the first equality we use that sheafification is a left adjoint to the inclusion of sheaves in presheaves. For the second equality we use that $f_ p$ is a left adjoint to $f_*$ on presheaves. We will return to this statement in the proof of Lemma 6.21.8.

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

commutes for all $V' \subset V \subset Y$ open.

In the literature we sometimes find this defined alternatively as in part (4) of Lemma 6.21.8 but as the lemma shows there is really no difference.

Lemma 6.21.8. Let $f : X \to Y$ be a continuous map. There are bijections between the following four sets

the set of maps $\mathcal{G} \to f_*\mathcal{F}$,

the set of maps $f^{-1}\mathcal{G} \to \mathcal{F}$,

the set of $f$-maps $\xi : \mathcal{G} \to \mathcal{F}$, and

the set of all collections of maps $\xi _{U, V} : \mathcal{G}(V) \to \mathcal{F}(U)$ for all $U \subset X$ and $V \subset Y$ open such that $f(U) \subset V$ compatible with all restriction maps,

functorially in $\mathcal{F} \in \mathop{\mathit{Sh}}\nolimits (X)$ and $\mathcal{G} \in \mathop{\mathit{Sh}}\nolimits (Y)$.

**Proof.**
A map of sheaves $a : \mathcal{G} \to f_*\mathcal{F}$ is by definition a rule which to each open $V$ of $Y$ assigns a map $a_ V : \mathcal{G}(V) \to f_*\mathcal{F}(V)$ and we have $f_*\mathcal{F}(V) = \mathcal{F}(f^{-1}(V))$. Thus at least the "data" corresponds exactly to what you need for an $f$-map $\xi $ from $\mathcal{G}$ to $\mathcal{F}$. To show that the sets (1) and (3) are in bijection we observe that $a$ is a map of sheaves if and only if corresponding family of maps $a_ V$ satisfy the condition in Definition 6.21.7.

Recall that $f^{-1}\mathcal{G}$ is the sheafification of $f_ p\mathcal{G}$. By the universal property of sheafification a map of sheaves $b : f^{-1}\mathcal{G} \to \mathcal{F}$ is the same thing as a map of presheaves $b_ p : f_ p\mathcal{G} \to \mathcal{F}$ where $f_ p$ is the functor defined earlier in the section. To give such a map $b_ p$ you need to specify for each open $U$ of $X$ a map

compatible with restriction mappings. We may and do view $b_{p, U}$ as a collection of maps $b_{p, U, V} : \mathcal{G}(V) \to \mathcal{F}(U)$ for all $V$ open in $Y$ with $f(U) \subset V$. These maps have to be compatible with all possible restriction mappings you can think of. In other words, we see that $b_ p$ corresponds to a collection of maps as in (4). Of course, conversely such a collection defines a map $b_ p$ and in turn a map $b : f^{-1}\mathcal{G} \to \mathcal{F}$.

To finish the proof of the lemma you have to show that by "forgetting structure" the rule that to a collection $\xi _{U, V}$ as in (4) associates the $f$-map $\xi $ with $\xi _ V = \xi _{f^{-1}(V), V}$ is bijective. To do this, if $\xi $ is a usual $f$-map then we just define $\tilde\xi _{U, V}$ to be the composition of $\xi _ V : \mathcal{G}(V) \to \mathcal{F}(f^{-1}(V))$ by the restruction map $\mathcal{F}(f^{-1}(V)) \to \mathcal{F}(U)$ which makes sense exactly because $f(U) \subset V$, i.e., $U \subset f^{-1}(V)$. This finishes the proof. $\square$

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

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

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

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

(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

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

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