I was wondering whether there was a notion of subcategory generated by a "part" of a fixed category $\mathcal{C}$. My thoughts started from the well known concept of the substructure generated by a subset of a given algebraic structure: e.g. the subgroup generated by a subset of a group, the sub-vector space generated by a subset of a vector space, the ideal generated by a subset of a ring and so on.....the spanned substructure is the minimal (w.r.t. inclusion) substructure which contains the subset or equivalently the intersection of all substructures containing the subset.
In the case of a category $\mathcal{C}$, consider the direct graph associated to $\mathcal{C}$ (maybe my notations are not standard), i.e. the graph with vertices given by objects of $\mathcal{C}$, edges given by morphisms, source and target functions given by domain and codomain of morphisms. Call it $G(\mathcal{C})$. Now consider a subgraph $X=(V_X,E_X)$ of $G_{\mathcal{C}}$. I define the subcategory generated by $X$ as the category $\langle X\rangle$ whose objects are vertices of $X$, and whose morphisms are $$Hom(\langle X\rangle):=\{f_1\circ\ldots\circ f_n: n\in\mathbb{N}, f_j\in E_X\}\cup \{id_V:V\in V_X\}$$ My questions:
1) Do you think that this makes sense?
2) If yes, do you think this notion is of some interest, or not?
3) If not, what is the correct notion, if exists? Where can I find some informations?
note: i'm not looking for the concept of free category generated by a graph. In that case, i've not a category $\mathcal{C}$ and i put an operation on the edges of the graph simply considering formal sequences of edges with concatenation, which is different i suppose..
Let $X\rightarrow G(\mathcal{C})$ be the inclusion of your subgraph. By adjunction, it defines $F(X)\rightarrow \mathcal{C}$ where $F(X)$ is the free category on the graph $X$. Then $\langle X\rangle$ is the image of that morphism. So I guess, it makes sense and is of interest. Note that, in general, one prefer to consider the essential image of a functor, and not just its strict image.