## 6.24 Continuous maps and sheaves of modules

The case of sheaves of modules is more complicated. The reason is that the natural setting for defining the pullback and pushforward functors, is the setting of ringed spaces, which we will define below. First we state a few obvious lemmas.

Lemma 6.24.1. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a presheaf of rings on $X$. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. There is a natural map of underlying presheaves of sets

$f_*\mathcal{O} \times f_*\mathcal{F} \longrightarrow f_*\mathcal{F}$

which turns $f_*\mathcal{F}$ into a presheaf of $f_*\mathcal{O}$-modules. This construction is functorial in $\mathcal{F}$.

Proof. Let $V \subset Y$ is open. We define the map of the lemma to be the map

$f_*\mathcal{O}(V) \times f_*\mathcal{F}(V) = \mathcal{O}(f^{-1}V) \times \mathcal{F}(f^{-1}V) \to \mathcal{F}(f^{-1}V) = f_*\mathcal{F}(V).$

Here the arrow in the middle is the multiplication map on $X$. We leave it to the reader to see this is compatible with restriction mappings and defines a structure of $f_*\mathcal{O}$-module on $f_*\mathcal{F}$. $\square$

Lemma 6.24.2. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a presheaf of rings on $Y$. Let $\mathcal{G}$ be a presheaf of $\mathcal{O}$-modules. There is a natural map of underlying presheaves of sets

$f_ p\mathcal{O} \times f_ p\mathcal{G} \longrightarrow f_ p\mathcal{G}$

which turns $f_ p\mathcal{G}$ into a presheaf of $f_ p\mathcal{O}$-modules. This construction is functorial in $\mathcal{G}$.

Proof. Let $U \subset X$ is open. We define the map of the lemma to be the map

\begin{eqnarray*} f_ p\mathcal{O}(U) \times f_ p\mathcal{G}(U) & = & \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{O}(V) \times \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{G}(V) \\ & = & \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} (\mathcal{O}(V)\times \mathcal{G}(V)) \\ & \to & \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{G}(V) \\ & = & f_ p\mathcal{G}(U). \end{eqnarray*}

Here the arrow in the middle is the multiplication map on $Y$. The second equality holds because directed colimits commute with finite limits, see Categories, Lemma 4.19.2. We leave it to the reader to see this is compatible with restriction mappings and defines a structure of $f_ p\mathcal{O}$-module on $f_ p\mathcal{G}$. $\square$

Let $f : X \to Y$ be a continuous map. Let $\mathcal{O}_ X$ be a presheaf of rings on $X$ and let $\mathcal{O}_ Y$ be a presheaf of rings on $Y$. So at the moment we have defined functors

\begin{eqnarray*} f_* : \textit{PMod}(\mathcal{O}_ X) & \longrightarrow & \textit{PMod}(f_*\mathcal{O}_ X) \\ f_ p : \textit{PMod}(\mathcal{O}_ Y) & \longrightarrow & \textit{PMod}(f_ p\mathcal{O}_ Y) \end{eqnarray*}

These satisfy some compatibilities as follows.

Lemma 6.24.3. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a presheaf of rings on $Y$. Let $\mathcal{G}$ be a presheaf of $\mathcal{O}$-modules. Let $\mathcal{F}$ be a presheaf of $f_ p\mathcal{O}$-modules. Then

$\mathop{Mor}\nolimits _{\textit{PMod}(f_ p\mathcal{O})}(f_ p\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{PMod}(\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.2 and 6.24.1, and we think of $f_*\mathcal{F}$ as an $\mathcal{O}$-module via the map $i_\mathcal {O} : \mathcal{O} \to f_*f_ p\mathcal{O}$ (defined first in the proof of Lemma 6.21.3).

Proof. Note that we have

$\mathop{Mor}\nolimits _{\textit{PAb}(X)}(f_ p\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{PAb}(Y)}(\mathcal{G}, f_*\mathcal{F}).$

according to Section 6.22. So what we have to prove is that under this correspondence, the subsets of module maps correspond. In addition, the correspondence is determined by the rule

$(\psi : f_ p\mathcal{G} \to \mathcal{F}) \longmapsto (f_*\psi \circ i_\mathcal {G} : \mathcal{G} \to f_* \mathcal{F})$

and in the other direction by the rule

$(\varphi : \mathcal{G} \to f_* \mathcal{F}) \longmapsto (c_\mathcal {F} \circ f_ p\varphi : f_ p\mathcal{G} \to \mathcal{F})$

where $i_\mathcal {G}$ and $c_\mathcal {F}$ are as in Section 6.22. Hence, using the functoriality of $f_*$ and $f_ p$ we see that it suffices to check that the maps $i_\mathcal {G} : \mathcal{G} \to f_* f_ p \mathcal{G}$ and $c_\mathcal {F} : f_ p f_* \mathcal{F} \to \mathcal{F}$ are compatible with module structures, which we leave to the reader. $\square$

Lemma 6.24.4. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a presheaf of rings on $X$. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a presheaf of $f_*\mathcal{O}$-modules. Then

$\mathop{Mor}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathcal{O} \otimes _{p, f_ pf_*\mathcal{O}} f_ p\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.2 and 6.24.1, and we use the map $c_\mathcal {O} : f_ pf_*\mathcal{O} \to \mathcal{O}$ in the definition of the tensor product.

Proof. This follows from the equalities

\begin{eqnarray*} \mathop{Mor}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathcal{O} \otimes _{p, f_ pf_*\mathcal{O}} f_ p\mathcal{G}, \mathcal{F}) & = & \mathop{Mor}\nolimits _{\textit{PMod}(f_ pf_*\mathcal{O})}( f_ p\mathcal{G}, \mathcal{F}_{f_ pf_*\mathcal{O}}) \\ & = & \mathop{Mor}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*(\mathcal{F}_{f_ pf_*\mathcal{O}})) \\ & = & \mathop{Mor}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

The first equality is Lemma 6.6.2. The second equality is Lemma 6.24.3. The third equality is given by the equality $f_*(\mathcal{F}_{f_ pf_*\mathcal{O}}) = f_*\mathcal{F}$ of abelian sheaves which is $f_*\mathcal{O}$-linear. Namely, $\text{id}_{f_*\mathcal{O}}$ corresponds to $c_\mathcal {O}$ under the adjunction described in the proof of Lemma 6.21.3 and thus $\text{id}_{f_*\mathcal{O}} = f_*c_\mathcal {O} \circ i_{f_*\mathcal{O}}$. $\square$

Lemma 6.24.5. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $X$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. The pushforward $f_*\mathcal{F}$, as defined in Lemma 6.24.1 is a sheaf of $f_*\mathcal{O}$-modules.

Proof. Obvious from the definition and Lemma 6.21.1. $\square$

Lemma 6.24.6. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $Y$. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}$-modules. There is a natural map of underlying presheaves of sets

$f^{-1}\mathcal{O} \times f^{-1}\mathcal{G} \longrightarrow f^{-1}\mathcal{G}$

which turns $f^{-1}\mathcal{G}$ into a sheaf of $f^{-1}\mathcal{O}$-modules.

Proof. Recall that $f^{-1}$ is defined as the composition of the functor $f_ p$ and sheafification. Thus the lemma is a combination of Lemma 6.24.2 and Lemma 6.20.1. $\square$

Let $f : X \to Y$ be a continuous map. Let $\mathcal{O}_ X$ be a sheaf of rings on $X$ and let $\mathcal{O}_ Y$ be a sheaf of rings on $Y$. So now we have defined functors

\begin{eqnarray*} f_* : \textit{Mod}(\mathcal{O}_ X) & \longrightarrow & \textit{Mod}(f_*\mathcal{O}_ X) \\ f^{-1} : \textit{Mod}(\mathcal{O}_ Y) & \longrightarrow & \textit{Mod}(f^{-1}\mathcal{O}_ Y) \end{eqnarray*}

These satisfy some compatibilities as follows.

Lemma 6.24.7. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $Y$. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}$-modules. Let $\mathcal{F}$ be a sheaf of $f^{-1}\mathcal{O}$-modules. Then

$\mathop{Mor}\nolimits _{\textit{Mod}(f^{-1}\mathcal{O})}(f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.6 and 6.24.5, and we think of $f_*\mathcal{F}$ as an $\mathcal{O}$-module by restriction via $\mathcal{O} \to f_*f^{-1}\mathcal{O}$.

Proof. Argue by the equalities

\begin{eqnarray*} \mathop{Mor}\nolimits _{\textit{Mod}(f^{-1}\mathcal{O})}(f^{-1}\mathcal{G}, \mathcal{F}) & = & \mathop{Mor}\nolimits _{\textit{Mod}(f_ p\mathcal{O})}(f_ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

where the second is Lemmas 6.24.3 and the first is by Lemma 6.20.1. $\square$

Lemma 6.24.8. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $X$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a sheaf of $f_*\mathcal{O}$-modules. Then

$\mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.6 and 6.24.5, and we use the canonical map $f^{-1}f_*\mathcal{O} \to \mathcal{O}$ in the definition of the tensor product.

Proof. This follows from the equalities

\begin{eqnarray*} \mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) & = & \mathop{Mor}\nolimits _{\textit{Mod}(f^{-1}f_*\mathcal{O})}( f^{-1}\mathcal{G}, \mathcal{F}_{f^{-1}f_*\mathcal{O}}) \\ & = & \mathop{Mor}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

which are a combination of Lemma 6.20.2 and 6.24.7. $\square$

Let $f : X \to Y$ be a continuous map. Let $\mathcal{O}_ X$ be a (pre)sheaf of rings on $X$ and let $\mathcal{O}_ Y$ be a (pre)sheaf of rings on $Y$. So at the moment we have defined functors

\begin{eqnarray*} f_* : \textit{PMod}(\mathcal{O}_ X) & \longrightarrow & \textit{PMod}(f_*\mathcal{O}_ X) \\ f_* : \textit{Mod}(\mathcal{O}_ X) & \longrightarrow & \textit{Mod}(f_*\mathcal{O}_ X) \\ f_ p : \textit{PMod}(\mathcal{O}_ Y) & \longrightarrow & \textit{PMod}(f_ p\mathcal{O}_ Y) \\ f^{-1} : \textit{Mod}(\mathcal{O}_ Y) & \longrightarrow & \textit{Mod}(f^{-1}\mathcal{O}_ Y) \end{eqnarray*}

Clearly, usually the pair of functors $(f_*, f^{-1})$ on sheaves of modules are not adjoint, because their target categories do not match. Namely, as we saw above, it works only if by some miracle the sheaves of rings $\mathcal{O}_ X, \mathcal{O}_ Y$ satisfy the relations $\mathcal{O}_ X = f^{-1}\mathcal{O}_ Y$ and $\mathcal{O}_ Y = f_*\mathcal{O}_ X$. This is almost never true in practice. We interrupt the discussion to define the correct notion of morphism for which a suitable adjoint pair of functors on sheaves of modules exists.

## Comments (4)

Comment #1072 by Comment on

In the proof of 6.24.3, one could mention that also the converse inclusion needs to be checked, and reduces to checking that $c_\mathcal{F}$ is compatible with the module structure.

In the last equality of 6.24.4 one uses the equality $f_*\mathcal{O}$-modules $f_*(\mathcal{F}_{f_pf_*\mathcal{O}})=f_*\mathcal{F}.$ It maybe helpful to point out to the reader, that the equality of the module structures follows, since $id_{f_*\mathcal{O}}$ corresponds to $c_\mathcal{O}$ under the adjunction described in the proof of Lemma 6.21.3 and thus $id_{f_*\mathcal{O}}=f_*c_\mathcal{O}\circ i_{f_*\mathcal{O}}$. One could make similar comments for the last equalities in the proofs of 6.24.8 and 6.26.2.

Comment #1076 by on

Hi, OK, I've fixed the lemma as you suggested. Thanks! You can find the change here.

As for Tag 008Z and 0096: we should fix these by referring in their proofs to Lemma Tag 008V and not reprove them as was done so far. It is a TODO.

Comment #1082 by Comment on

Thanks for your great work! I think there is still a typo in the proof of Lemma 6.24.4. It should say $f_*\mathcal{O}$-linear instead of $\mathcal{O}$-linear in the sentence:

The third equality is given by the equality $f_*(\mathcal{F}_{f_p f_*\mathcal{O}}) = f_*\mathcal{F}$ of abelian sheaves which is $\mathcal{O}$-linear.

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