Remark 22.25.7. Let $R$ be a ring. Let $\mathcal{D}$ be an $R$-linear category endowed with a collection of $R$-linear functors $[n] : \mathcal{D} \to \mathcal{D}$, $x \mapsto x[n]$ indexed by $n \in \mathbf{Z}$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}_\mathcal {D}$ (equality as functors). This allows us to construct a graded category $\mathcal{D}^{gr}$ over $R$ with the same objects of $\mathcal{D}$ setting

for $x, y$ in $\mathcal{D}$. Observe that $(\mathcal{D}^{gr})^0 = \mathcal{D}$ (see Definition 22.25.3). Moreover, the graded category $\mathcal{D}^{gr}$ inherits $R$-linear graded functors $[n]$ satisfying $[n] \circ [m] = [n + m]$ and $[0] = \text{id}_{\mathcal{D}^{gr}}$ with the property that

as graded $R$-modules compatible with composition of morphisms.

Conversely, suppose given a graded category $\mathcal{A}$ over $R$ endowed with a collection of $R$-linear graded functors $[n]$ satisfying $[n] \circ [m] = [n + m]$ and $[0] = \text{id}_\mathcal {A}$ which are moreover equipped with isomorphisms

as graded $R$-modules compatible with composition of morphisms. Then the reader easily shows that $\mathcal{A} = (\mathcal{A}^0)^{gr}$.

Here are two examples of the relationship $\mathcal{D} \leftrightarrow \mathcal{A}$ we established above:

Let $\mathcal{B}$ be an additive category. If $\mathcal{D} = \text{Gr}(\mathcal{B})$, then $\mathcal{A} = \text{Gr}^{gr}(\mathcal{B})$ as in Example 22.25.5.

If $A$ is a graded ring and $\mathcal{D} = \text{Mod}_ A$ is the category of graded right $A$-modules, then $\mathcal{A} = \text{Mod}^{gr}_ A$, see Example 22.25.6.

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