The Stacks project

Lemma 18.36.2. Let $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ be a ringed topos. Let $p$ be a point of the topos $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$.

  1. The functor $p_* : \textit{Mod}(\mathcal{O}_ p) \to \textit{Mod}(\mathcal{O})$, $M \mapsto p_*M$ is exact.

  2. The canonical surjection $p^{-1}p_*M \to M$ is $\mathcal{O}_ p$-linear.

  3. The functorial direct sum decomposition $p^{-1}p_*M = M \oplus I(M)$ of Lemma 18.36.1 is not $\mathcal{O}_ p$-linear in general.

Proof. Part (1) and surjectivity in (2) follow immediately from the corresponding result for abelian sheaves in Lemma 18.36.1. Since $p^{-1}\mathcal{O} = \mathcal{O}_ p$ we have $p^{-1} = p^*$ and hence $p^{-1}p_*M \to M$ is the same as the counit $p^*p_*M \to M$ of the adjunction for modules, whence linear.

Proof of (3). Suppose that $G$ is a group. Consider the topos $G\textit{-Sets} = \mathop{\mathit{Sh}}\nolimits (\mathcal{T}_ G)$ and the point $p : \textit{Sets} \to G\textit{-Sets}$. See Sites, Section 7.9 and Example 7.33.7. Here $p^{-1}$ is the functor forgetting about the $G$-action. And $p_*$ is the right adjoint of the forgetful functor, sending $M$ to $\text{Map}(G, M)$. The maps in the direct sum decomposition are the maps

\[ M \to \text{Map}(G, M) \to M \]

where the first sends $m \in M$ to the constant map with value $m$ and where the second map is evaluation at the identity element $1$ of $G$. Next, suppose that $R$ is a ring endowed with an action of $G$. This determines a sheaf of rings $\mathcal{O}$ on $\mathcal{T}_ G$. The category of $\mathcal{O}$-modules is the category of $R$-modules $M$ endowed with an action of $G$ compatible with the action on $R$. The $R$-module structure on $\text{Map}(G, M)$ is given by

\[ ( r f ) (\sigma ) = \sigma (r) f(\sigma ) \]

for $r \in R$ and $f \in \text{Map}(G, M)$. This is true because it is the unique $G$-invariant $R$-module strucure compatible with evaluation at $1$. The reader observes that in general the image of $M \to \text{Map}(G, M)$ is not an $R$-submodule (for example take $M = R$ and assume the $G$-action is nontrivial), which concludes the proof. $\square$


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