No commuting diagrams in a free category?

Question

Is it correct that given a diagram $D$, that in the free category on $D$ there are no commuting diagrams consisting solely of morphisms that are also in $D$, and that don’t have identity morphisms in them?

Answer 1

No. For instance, the empty diagram commutes in any category! So does a diagram $A\to B$ with just a single morphism (between different vertices). Or, you could have a diagram like $$\require{AMScd} \begin{CD} A @>{f}>> B\\ @V{f}VV @VV{g}V \\ B @>{g}>> C \end{CD}$$ which commutes because some of its arrows happen to be labelled with the same morphism.

You could probably come up with some combinatorial condition that will guarantee that a diagram doesn't commute, but I think it's much more fruitful to just understand the explicit characterization of morphisms in the free category. Namely, if $D$ is a directed graph and $F$ is the free category on $D$, then a morphism in $F$ between two vertices $x,y$ of $D$ is just a directed path in $D$ from $x$ to $y$, and composition of morphisms is concatenation of paths. So, for instance, a composition of two edges of $D$ can never equal another edge of $D$, since the composition is a path of length $2$.