The paragraph reads: Let $T$ be a topos, and let $\Lambda$ be a ring in $T$. Denote by $\operatorname{Mod}_{\Lambda}$ the category of $\Lambda$-modules in $T$. My question is how is the category of $\Lambda$-modules defined?

The text defined a sheaf of abelian group(rings) in $T$ as a sheaf $A\in T$, together with morphisms of sheaves of sets $m: A\times A\to A$ and $e: \{*\}\to A$, satisfying the corresponding axioms. I assume a $\Lambda$-module $M$ would be a sheaf of abelian groups, along with an action of $\Lambda$ on it, but I don't see how to define the action of $\Lambda$ on $M$ in this general setting.


1 Answer 1


The definition works in every category $\mathcal{C}$ with finite products. We define a $\Lambda$-module in $\mathcal{C}$ to be an abelian group $A$ in $\mathcal{C}$ together with a morphism $\cdot : \Lambda \times A \to A$ such that the diagrams commute which correspond to the module axioms. For example, $1 \cdot a = a$ means that $1 \times A \xrightarrow{1 \times A} \Lambda \times A \xrightarrow{\cdot} A$ is equal to $p_2$. Every module axiom (and also every ring axiom) can also be written down more conveniently with generalized elements. The previous axiom really just says that $1 \cdot a = a$ holds for all generalized elements $a \in A$. If $\mathcal{C}$ is the topos of sheaves on a space, this means that $1 \cdot a = a$ holds for all local sections $a$ of $A$.

  • $\begingroup$ Another way of putting the construction using generalized elements would be in terms of Yoneda's lemma: $\cdot : \Lambda \times A \to A$ corresponds to a morphism of functors $\operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},A) \to \operatorname{Hom}({-},A)$, and similarly for the ring operations on $\Lambda$ and group operations on $A$. So then for example, the two sides of distributivity $(a + b) x = ax + bx$ give morphisms $\operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},A)\to\operatorname{Hom}({-},A)$, which... $\endgroup$ Jan 13, 2023 at 21:47
  • $\begingroup$ then correspond to two morphisms $\Lambda\times\Lambda\times A\to A$, and you then assert the equality of those two morphisms. $\endgroup$ Jan 13, 2023 at 21:48
  • $\begingroup$ Or, given that we're working in a topos, we could just cast the identities as statements in the internal language of a topos, e.g. $(\forall a:\Lambda)(\forall x,y:A) a(x+y)=ax+ay$, and asserting that the interpretation of that statement is equal to $\top : \operatorname{Hom}(1, \Omega)$ would turn out to be exactly what we need. $\endgroup$ Jan 13, 2023 at 21:51
  • $\begingroup$ That's right. Thx $\endgroup$ Jan 13, 2023 at 22:53

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