>**Puzzle 121.** Let \\(\mathbf{N}\\) be the free category on this graph:

>

>It has one morphism for each natural number. What are all the functors \\(F: \mathbf{N} \to \mathbf{N}\\)? What happens when we compose them?

Such a functor, \\(F\\), must map 0 (\\(id_\mathbf{N}\\)) to 0,

\\[
F(id_\mathbf{N}) = id_\mathbf{N} = 0,
\\]

and it must send the non-identity \\(s\\) to either \\(id_\mathbf{N}\\), \\(s\\), or \\(s\\) repeated a certain finite number of times (\\(s\circ s\circ \cdots \circ s \\)),

\\[
F(s) =id_\mathbf{N}
\\]

\\[
\Leftrightarrow \\\\ F(s\circ s) = F(s)\circ F(s) = id_\mathbf{N} \circ id_\mathbf{N} = 0 + 0,
\\]

\\[
\Leftrightarrow \\\\ F(\underbrace{s\circ s\circ \cdots \circ s}\_{n \text{ times }}) \\\\
= \underbrace{F(s)\circ F(s)\circ \cdots \circ F(s)}\_{n \text{ times }} \\\\
=\underbrace{id_\mathbf{N} \circ id_\mathbf{N} \circ \cdots \circ id_\mathbf{N} }\_{n \text{ times }} \\\\
=\underbrace{0 + 0 + \cdots 0}\_{n \text{ times }} = 0
\\]

and,
\\[
F(s) =s = 1
\\]
\\[
\Leftrightarrow \\\\ F(s\circ s) = F(s)\circ F(s) =s \circ s =1+1 ,
\\]
\\[
\Leftrightarrow \\\\
F(\underbrace{s\circ s\circ \cdots \circ s}\_{n \text{ times }}) \\\\
= \underbrace{F(s)\circ F(s)\circ \cdots \circ F(s)}\_{n \text{ times }} \\\\
=\underbrace{s\circ s\circ \cdots \circ s}\_{n \text{ times }} \\\\
=\underbrace{1 + 1 + \cdots 1}\_{n \text{ times }} = n
\\]

which makes such an \\(F\\) the identity functor and,

\\[
F(s) = \underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }} = m
\\]
\\[
\Leftrightarrow \\\\
F(s\circ s) = F(s)\circ F(s) = \underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }} \circ \underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }} = m+m = 2m,
\\]
\\[
\Leftrightarrow \\\\
F(\underbrace{s\circ s\circ \cdots \circ s}\_{n \text{ times }}) \\\\
= \underbrace{F(s)\circ F(s)\circ \cdots \circ F(s)}\_{n \text{ times }} \\\\
=\underbrace{
\underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }}\circ
\underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }}\circ \cdots \circ
\underbrace{s\circ s\circ \cdots \circ s}\_{m \text{ times }}}\_{n \text{ times }} \\\\
=\underbrace{m + m + \cdots m}\_{n \text{ times }} = m*n
\\]

which would imply that functors \\(F: \mathbf{N} \to \mathbf{N}\\) are multiplication.