>Here's an example. Take \\(\mathcal{C} = \mathbf{Free}(G)\\) and look at the morphisms \\(\textrm{WorksIn} : \textrm{Employee} \to \textrm{Department}\\) and

\\(\textrm{DepartmentName} : \textrm{Department} \to \textrm{String}\\). Then choose a functor \\(F : \mathcal{C} \to \mathbf{Set}\\). We must have

>\[ F(\textrm{DepartmentName} \circ \textrm{WorksIn}) = F(\textrm{DepartmentName}) \circ F(\textrm{WorksIn}) .\]

>**Puzzle 109.** What does this mean, in practical terms?

In practical terms, it means that if we take any employee, finding their department they actually work in, *and then* finding the name of the department they work in (maybe you want to report them...), is the same taking the same employee, and directly finding out the department name they work in (as one operation instead of two).

\\(\textrm{DepartmentName} : \textrm{Department} \to \textrm{String}\\). Then choose a functor \\(F : \mathcal{C} \to \mathbf{Set}\\). We must have

>\[ F(\textrm{DepartmentName} \circ \textrm{WorksIn}) = F(\textrm{DepartmentName}) \circ F(\textrm{WorksIn}) .\]

>**Puzzle 109.** What does this mean, in practical terms?

In practical terms, it means that if we take any employee, finding their department they actually work in, *and then* finding the name of the department they work in (maybe you want to report them...), is the same taking the same employee, and directly finding out the department name they work in (as one operation instead of two).