Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 1.2.6.11. Let $\operatorname{\mathcal{C}}$ be a category. The following conditions on $\operatorname{\mathcal{C}}$ are equivalent:

$(a)$

The category $\operatorname{\mathcal{C}}$ is free. That is, there exists a directed graph $G$ and an isomorphism of categories $\operatorname{\mathcal{C}}\simeq \operatorname{Path}[G]$.

$(b)$

The functor $F: \operatorname{Path}[ \mathrm{Gr}_0(\operatorname{\mathcal{C}})] \rightarrow \operatorname{\mathcal{C}}$ an isomorphism of categories.

$(c)$

The functor $F: \operatorname{Path}[ \mathrm{Gr}_0(\operatorname{\mathcal{C}})] \rightarrow \operatorname{\mathcal{C}}$ is an equivalence of categories.

$(d)$

The functor $F: \operatorname{Path}[ \mathrm{Gr}_0(\operatorname{\mathcal{C}})] \rightarrow \operatorname{\mathcal{C}}$ is fully faithful.

$(e)$

Every morphism $f$ in $\operatorname{\mathcal{C}}$ admits a unique factorization $f = f_{n} \circ f_{n-1} \circ \cdots \circ f_1$, where each $f_{i}$ is an indecomposable morphism of $\operatorname{\mathcal{C}}$.

Proof. The functor $F$ is bijective on objects, which shows that $(b) \Leftrightarrow (c) \Leftrightarrow (d)$. The equivalence of $(d)$ and $(e)$ follows from the definition of morphisms in the path category $\operatorname{Path}[ \mathrm{Gr}_0(\operatorname{\mathcal{C}})]$. The implication $(b) \Rightarrow (a)$ is immediate, and the converse follows from Example 1.2.6.9. $\square$