ct.category theory - Parametrized natural numbers object.

I give two examples of categories with finite products and simple NNO. In the first example the simple NNO is also a parameterized NNO, while in the second example it is not. Although it is difficult to understand your question, I believe the examples should clarify matters.

First, consider the category $mathcal{C}$ whose objects are the finite powers of $mathbb{N}$, namely $mathbb{N}^0$, $mathbb{N}^1$, $mathbb{N}^2$, ... and morphisms are set-theoretic functions $f : mathbb{N}^k to mathbb{N}^m$. This category clearly has finite products, is not cartesian-closed because there are too many morhisms $mathbb{N} to mathbb{N}$, and it has a parameterized NNO, namely the obvious one.

Second, consider the category $mathcal{D}$ whose objects are the finite powers of $mathbb{N}$, like before, and whose morphisms are as follows:

1. Morphisms $mathbb{N}^k to mathbb{N}^m$ with $m neq 1$ are all set-theoretic functions.


2. Morphisms $mathbb{N}^0 to mathbb{N}^1$ are all set-theoretic functions, i.e., for each natural number there is one.


3. Morphisms $mathbb{N}^k to mathbb{N}^1$ with $k neq 0$ are all set-theoretic functions $f : mathbb{N}^k to mathbb{N}$ for which there exists a projection $pi_j : mathbb{N}^k to mathbb{N}$ and $g : mathbb{N} to mathbb{N}$ such that $f = g circ pi_j$.

In other words, in $mathcal{D}$ every function into $mathbb{N}$ depends on only one of its parameters (exercise: prove that these are closed under composition.) The category $mathcal{D}$ has finite products and a simple NNO, namely the obvious one, but no parameterized NNO. If it did, we could construct addition ${+} : mathbb{N}^2 to mathbb{N}$ as a morphism in the category.