Let $Q$ be a quiver defined as a covariant functor from this category $\chi$: $$\left\{Q_1 \overset{h}{\underset{t}{\Large\rightrightarrows}} Q_0 \right\}\overset{\Gamma}\longrightarrow\mathsf {Set}$$ to $\mathsf{Set}$. This is all great and well as a definition of a quiver but I am finding it really difficult to define or think about the free category of this quiver. The way I would construct $F(\Gamma)$ for some quiver $\Gamma$ would be as follows:
- Take $\mathrm{Ob}(\Gamma)$ to be $Q_0$.
- For each $e \in Q_1$ add it to $\hom(h(e), t(e))$
- Categorify this by adding in composites to the homs and identities to the objects
This definition of a free category doesn't feel very natural to me, not to mention the difficulties arising from this construction basically being an algorithm.
What is the natural/categorical way to talk about the free category of a quiver defined like this?
It's not clear to me what you find unsatisfying, but perhaps the following helps. Define a path (of length $n$) to be a pair of tuples $$((c_0,\dots,c_n),(f_1,\dots,f_n))\in\Gamma(Q_0)^{n+1}\times\Gamma(Q_1)^n$$ such that $h(f_i)=c_{i-1}$ and $t(f_i)=c_i$ for $i=1,\dots,n$. Intuitively, this object represents the formal composition of the arrows $f_i$.
You can then describe the free category on $\Gamma$ very explicitly as follows. The set of objects is $\Gamma(Q_0)$. Given $a,b\in \Gamma(Q_0)$, $\operatorname{Hom}(a,b)$ is the set of paths $((c_0,\dots,c_n),(f_1,\dots,f_n))$ such that $c_0=a$ and $c_n=b$. Composition is then just concatenation of paths: given $$((a,c_1,\dots,c_{n-1},b),(f_1,\dots,f_n))\in\operatorname{Hom}(a,b)$$ and $$((b,c_1',\dots,c_{m-1}',c),(f'_1,\dots,f'_m))\in\operatorname{Hom}(b,c),$$ their composition is the path $$((a,c_1\dots,c_{n-1},b,c_1',\dots,c_{m-1}',c),(f_1,\dots,f_n,f_1',\dots,f_m'))\in\operatorname{Hom}(a,c).$$ The identity morphism from $a\in\Gamma(Q_0)$ to itself is just the unique path of length $0$ from $a$ to itself: $((a),\epsilon)$ where $\epsilon$ is the empty tuple.