**Puzzle 131.** How many functors are there from \\(\mathbf{2}\\) to \\(\mathbf{3}\\), and how many natural transformations are there between all these functors? Again, it may help to draw a graph.

There are three functors \\(F\\), \\(F'\\), and \\(F''\\) mapping \\(f_1\\) in \\(\mathbf{2}\\) to morphisms of lengths 0, 1 and 2 sending \\(v_1\\) in \\(\mathbf{2}\\) to \\(v_1\\) in \\(\mathbf{3}\\). Similarly two functors \\(G\\) and \\(G'\\) sending \\(f_1\\) to morphisms of length 0 or 1 starting at \\(v_2\\). Last a functor \\(H\\) sending \\(f_1\\) to the identity on \\(v_3\\).

So six functors.

I drew a graph of natural transformations between them, and avoiding the urge to invent new backwards morphisms this time, came up with \\(F \Rightarrow F' \Rightarrow F''\\), and \\(G \Rightarrow G' \Rightarrow H\\), with further transformations \\(F' \Rightarrow G\\) and \\(F'' \Rightarrow G'\\).

The total number of natural transformations will be the count of all paths on this graph:

- 6 of length 0 (the identity transformations)
- 6 of length 1 (the transformation arrows on the graph)
- 6 of length 2 (three via F'', three via G)
- 4 of length 3 (two via F'', two via G)
- 2 of length 4 (one via F'', one via G)

That's a total of 24 natural transformations between functors from \\(\mathbf{2}\\) to \\(\mathbf{3}\\).