### Operadification II

Underlying the concept of natural number object is the basic recursion theorem. The composition of the arrows

$f \circ q: 1 \rightarrow A \rightarrow A$

in Set is just the evaluation $f(q)$. This arrow $f(q): 1 \rightarrow A$ can itself be used as input for the same diagram, by appending another copy of $f$ to the right, to obtain the arrow $f(f(q))$. That is, the natural number object commuting diagram extends indefinitely to the right by appending extra copies of the successor and the function $f$. Once the comparison arrow $u: N \rightarrow A$ assigns zero to $q$, it follows that it must assign $f(q)$ to 1, $f(f(q))$ to 2, and so on.

Thus the definition of recursion, as a possibly infinite process, demands the full set $N$ rather than some finite ordinal set. But for periodic recursive functions, satisfying $f(f(f(\cdots(q)))) = f(q)$ for $n + 1$ brackets on the left hand side, modular arithmetic using the set n acts as a universal diagram. For example, if $f$ represents rotation by an $n$-th root of unity, then it is periodic in this sense.

$f \circ q: 1 \rightarrow A \rightarrow A$

in Set is just the evaluation $f(q)$. This arrow $f(q): 1 \rightarrow A$ can itself be used as input for the same diagram, by appending another copy of $f$ to the right, to obtain the arrow $f(f(q))$. That is, the natural number object commuting diagram extends indefinitely to the right by appending extra copies of the successor and the function $f$. Once the comparison arrow $u: N \rightarrow A$ assigns zero to $q$, it follows that it must assign $f(q)$ to 1, $f(f(q))$ to 2, and so on.

Thus the definition of recursion, as a possibly infinite process, demands the full set $N$ rather than some finite ordinal set. But for periodic recursive functions, satisfying $f(f(f(\cdots(q)))) = f(q)$ for $n + 1$ brackets on the left hand side, modular arithmetic using the set n acts as a universal diagram. For example, if $f$ represents rotation by an $n$-th root of unity, then it is periodic in this sense.

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