Modules over a category ring

Let $𝖢$ be a category with finite $O b (𝖢)$ . Then a module over the the category ring $𝕂 𝖢$ is equivalent to a functor $𝖢 \to 𝖵 𝖾 𝖼 𝗍 𝕂$ , #m/thm/rep and we have an equivalence of categories¹

𝕂 [𝖢] 𝖬 𝗈 𝖽 ≃ 𝖵 𝖾 𝖼 𝗍 𝕂 𝖢 .

Proof

Let $𝐹 : 𝖢 \to 𝖵 𝖾 𝖼 𝗍 𝕂$ be a functor, Let $(𝑉 𝐹) 𝑋 = 𝐹 𝑋$ and thus construct the $O b (𝖢)$ -graded vector space

𝑉 𝐹 = ⨁ 𝑋 \in O b (𝖢) (𝑉 𝐹) 𝑋

and for a morphism $𝑓 \in 𝖢$ and homogenous vector $𝑣 \in (𝑉 𝐹) 𝑋$ define

𝑓 ⊙ 𝑣 = {(𝐹 𝑓) 𝑣 𝑋 = d o m 𝑓 0 𝑋 \neq d o m 𝑓

and extend this definition linearly first for nonhomogenous vectors and then general $𝑓 \in 𝕂 [𝖢]$ . Clearly this defines a $𝕂 [𝖢]$ -module.

Now suppose $𝜑 \in 𝖵 𝖾 𝖼 𝗍 𝕂 𝖢 (𝐹, 𝐺)$ is a natural transformation with components $𝜑 𝑋 : (𝑉 𝐹) 𝑋 \to (𝑉 𝐺) 𝑋$ . Then

(𝑉 𝜑) = ⨁ 𝑋 \in O b (𝖢) 𝜑 𝑋 : 𝑉 𝐹 \to 𝑉 𝐺

defines an $O b (𝖢)$ -graded linear map. Moreover, by naturality of $𝜑$ , for a morphism $𝑓 \in 𝖢 (𝑋, 𝑌)$ and homogenous vector $𝑣 \in (𝑉 𝐹) 𝑋$

(𝑉 𝜑) (𝑓 ⊙ 𝑣) = 𝜑 𝑌 (𝐹 𝑓) 𝑣 = (𝐺 𝑓) 𝜑 𝑋 𝑣 = 𝑓 ⊙ (𝑉 𝜑) 𝑣

so by linearity $𝑉 𝜑$ is a $𝕂 [𝖢]$ -module isomorphism. Therefore $𝑉 : 𝖵 𝖾 𝖼 𝗍 𝕂 𝖢 \to 𝕂 [𝖢] 𝖬 𝗈 𝖽$ is a functor.

Conversely, suppose $𝑉$ is a $𝕂 [𝖢]$ -module. We define a functor $𝑀 𝑉 : 𝖢 \to 𝖵 𝖾 𝖼 𝗍 𝕂$ as follows:

$(𝑀 𝑉) 𝑋 = 1 𝑋 ⊙ 𝑉$ for $𝑋 \in O b (𝖢)$ ;
$((𝑀 𝑉) 𝑓) 𝑣 = 𝑓 ⊙ 𝑣$ for $𝑓 \in 𝖢 (𝑋, 𝑌)$ and $𝑣 \in (𝑀 𝑉) 𝑋$ .

Now suppose $𝜑 : 𝑉 \to 𝑊$ is a $𝕂 [𝖢]$ -module homomorphism. We define a transformation with components

(𝑀 𝜑) 𝑋 : (𝑀 𝑉) 𝑋 \to (𝑀 𝑊) 𝑋 𝑣 \mapsto 𝜑 𝑣

which is well-defined since $𝜑$ is $O b (𝖢)$ -graded. Moreover, for any $𝑓 \in 𝖢 (𝑋, 𝑌)$ and $𝑣 \in 𝑀 (𝑉) 𝑋$

((𝑀 𝑊) 𝑓) (𝑀 𝜑) 𝑋 𝑣 = 𝑓 ⊙ 𝜑 𝑣 = 𝜑 (𝑓 ⊙ 𝑣) = (𝑀 𝜑) 𝑌 ((𝑀 𝑉) 𝑓)

so the following diagram commutes

whence $𝑀 𝜑$ is natural and $𝑀 : 𝕂 [𝖢] 𝖬 𝗈 𝖽 \to 𝖵 𝖾 𝖼 𝗍 𝕂 𝖢$ is a functor.

It is not difficult to see the natural equivalences required to make this an equivalence.

#state/tidy | #lang/en | #SemBr

assuming the Axiom of Choice. ↩

Modules over a category ring

Footnotes