**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}\$$.