## 18.36 Stalks of modules

We have to be a bit careful when taking stalks at points, since the colimit defining a stalk (see Sites, Equation 7.32.1.1) may not be filtered^{1}. On the other hand, by definition of a point of a site the stalk functor is exact and commutes with arbitrary colimits. In other words, it behaves exactly as if the colimit were filtered.

Lemma 18.36.1. Let $\mathcal{C}$ be a site. Let $p$ be a point of $\mathcal{C}$.

We have $(\mathcal{F}^\# )_ p = \mathcal{F}_ p$ for any presheaf of sets on $\mathcal{C}$.

The stalk functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact (see Categories, Definition 4.23.1) and commutes with arbitrary colimits.

The stalk functor $\textit{PSh}(\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact (see Categories, Definition 4.23.1) and commutes with arbitrary colimits.

**Proof.**
By Sites, Lemma 7.32.5 we have (1). By Sites, Lemmas 7.32.4 we see that $\textit{PSh}(\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is a left adjoint, and by Sites, Lemma 7.32.5 we see the same thing for $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$. Hence the stalk functor commutes with arbitrary colimits (see Categories, Lemma 4.24.5). It follows from the definition of a point of a site, see Sites, Definition 7.32.2 that $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact. Since sheafification is exact (Sites, Lemma 7.10.14) it follows that $\textit{PSh}(\mathcal{C}) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact.
$\square$

In particular, since the stalk functor $\mathcal{F} \mapsto \mathcal{F}_ p$ on presheaves commutes with all finite limits and colimits we may apply the reasoning of the proof of Sites, Proposition 7.44.3. The result of such an argument is that if $\mathcal{F}$ is a (pre)sheaf of algebraic structures listed in Sites, Proposition 7.44.3 then the stalk $\mathcal{F}_ p$ is naturally an algebraic structure of the same kind. Let us explain this in detail when $\mathcal{F}$ is an abelian presheaf. In this case the addition map $+ : \mathcal{F} \times \mathcal{F} \to \mathcal{F}$ induces a map

\[ + : \mathcal{F}_ p \times \mathcal{F}_ p = (\mathcal{F} \times \mathcal{F})_ p \longrightarrow \mathcal{F}_ p \]

where the equal sign uses that stalk functor on presheaves of sets commutes with finite limits. This defines a group structure on the stalk $\mathcal{F}_ p$. In this way we obtain our stalk functor

\[ \textit{PAb}(\mathcal{C}) \longrightarrow \textit{Ab}, \quad \mathcal{F} \longmapsto \mathcal{F}_ p \]

By construction the underlying set of $\mathcal{F}_ p$ is the stalk of the underlying presheaf of sets. This also defines our stalk functor for sheaves of abelian groups by precomposing with the inclusion $\textit{Ab}(\mathcal{C}) \subset \textit{PAb}(\mathcal{C})$.

Lemma 18.36.2. Let $\mathcal{C}$ be a site. Let $p$ be a point of $\mathcal{C}$.

The functor $\textit{Ab}(\mathcal{C}) \to \textit{Ab}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact.

The stalk functor $\textit{PAb}(\mathcal{C}) \to \textit{Ab}$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact.

For $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\textit{PAb}(\mathcal{C}))$ we have $\mathcal{F}_ p = \mathcal{F}^\# _ p$.

**Proof.**
This is formal from the results of Lemma 18.36.1 and the construction of the stalk functor above.
$\square$

Next, we turn to the case of sheaves of modules. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. (It suffices for the discussion that $\mathcal{O}$ be a presheaf of rings.) Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. Let $p$ be a point of $\mathcal{C}$. In this case we get a map

\[ \cdot : \mathcal{O}_ p \times \mathcal{O}_ p = (\mathcal{O} \times \mathcal{O})_ p \longrightarrow \mathcal{O}_ p \]

which is the stalk of the multiplication map and

\[ \cdot : \mathcal{O}_ p \times \mathcal{F}_ p = (\mathcal{O} \times \mathcal{F})_ p \longrightarrow \mathcal{F}_ p \]

which is the stalk of the multiplication map. We omit the verification that this defines a ring structure on $\mathcal{O}_ p$ and an $\mathcal{O}_ p$-module structure on $\mathcal{F}_ p$. In this way we obtain a functor

\[ \textit{PMod}(\mathcal{O}) \longrightarrow \textit{Mod}(\mathcal{O}_ p), \quad \mathcal{F} \longmapsto \mathcal{F}_ p \]

By construction the underlying set of $\mathcal{F}_ p$ is the stalk of the underlying presheaf of sets. This also defines our stalk functor for sheaves of $\mathcal{O}$-modules by precomposing with the inclusion $\textit{Mod}(\mathcal{O}) \subset \textit{PMod}(\mathcal{O})$.

Lemma 18.36.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $p$ be a point of $\mathcal{C}$.

The functor $\textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_ p)$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact.

The stalk functor $\textit{PMod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_ p)$, $\mathcal{F} \mapsto \mathcal{F}_ p$ is exact.

For $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\textit{PMod}(\mathcal{O}))$ we have $\mathcal{F}_ p = \mathcal{F}^\# _ p$.

**Proof.**
This is formal from the results of Lemma 18.36.2, the construction of the stalk functor above, and Lemma 18.14.1.
$\square$

Lemma 18.36.4. Let $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi or ringed sites. Let $p$ be a point of $\mathcal{C}$ or $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ and set $q = f \circ p$. Then

\[ (f^*\mathcal{F})_ p = \mathcal{F}_ q \otimes _{\mathcal{O}_{\mathcal{D}, q}} \mathcal{O}_{\mathcal{C}, p} \]

for any $\mathcal{O}_\mathcal {D}$-module $\mathcal{F}$.

**Proof.**
We have

\[ f^*\mathcal{F} = f^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}_\mathcal {C} \]

by definition. Since taking stalks at $p$ (i.e., applying $p^{-1}$) commutes with $\otimes $ by Lemma 18.26.2 we win by the relation between the stalk of pullbacks at $p$ and stalks at $q$ explained in Sites, Lemma 7.34.2 or Sites, Lemma 7.34.3.
$\square$

## Comments (2)

Comment #3408 by Mike Paluch on

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