In this and subsequent pages, we will often abuse the terminology and call the structure a graph, even though it is a digraph.
The graph allows loops and multiple edges. You can represent a morphism from an object into itself, or several different morphisms between two objects.
You may recall the abstract definition of a category, which is based on morphisms alone. An object is merely an identity morphism. This definition inspires an additional constraint on a diagram: each object includes the identity morphism on itself. Every vertex has an arc that loops around and runs back into itself. Let's assume this additional constraint is in place.
As a corollary, there is at most one arc directly connecting any two vertices. Multiple edges have just disappeared. And a literal interpretation means there is at most one arc joining a vertex to itself, and we already have that; it's the identity map. There are no other loops.
If a cycle starts and ends at V, the composition of those morphisms has to produce the morphism on V, which is the identity map. If U and V are connected by antiparallel arcs, that forms a cycle of length 2. The composition is the identity map, the morphisms are inverses, and the points are equivalent.
Note that an endomorphism can be included in a commutative diagram; you just need to vertices that represent the same object. Draw an arc between them, and it won't comflict with the identity arc, which is a loop at either vertex. If you want to depict multiple morphisms between two objects, use several vertices to represent those objects.
The trick is defining the new morphisms. Within C1, let f take A to B, and let g take C to D. Draw a commutative diagram like this.
A f B x y C g D
x and y are morphisms in C1, yet to be determined. Arrows point down and to the right. We are only interested in commutative diagrams. Thus y(f(A)) = g(x(A)). Each ordered pair of morphisms x,y that makes the diagram commute becomes a morphism from f to g in C2.
Let's verify the category axioms, starting with composition. Stack two squares on top of each other, like this.
* f * w y * g * x z * h *
If the two squares form commutative diagrams, use the properties of associativity in C1 to write wxh = wgz = fyz. In other words, the entire rectangle commutes, and wx,yz is a valid morphism from f to h. Let this be the composition of w,y and x,z.
If I means identity map in C1, show that I,I is the identity morphism for any function f. Remember, you need to show that I,I joins with any other morphism to leave it unchanged.
* f * I I * f * x y * g *
Since morphism composition in C2 follows directly from morphism composition in C1, associativity is inherited. Therefore C2 forms a category.