Let $N$ be a set containing an element $1$ and $\sigma: N \to N$ an injective function satisfying the following two properties:
$\tag 1 1 \notin \sigma(N)$
$\tag 2 (\forall M \subset N) \;\text{If } [\; 1 \in M \land (\sigma(M) \subset M) \;] \text{ Then } M = N$
We call $(N, 1, \sigma)$ a Peano system.
The set of all injective functions on $N$ form a semigroup under composition.
Let $\mathcal C$ denote the set of all injective functions on $N$ that commute with $\sigma$. It is easy to see that $\mathcal C$ is a commutative semigroup containing the identity transformation.
Theorem 1: If $\mu,\nu \in \mathcal C$ and $\mu(1) = \nu(1)$ then $\mu = \nu$.
Proof
Let $M$ be the set of all elements in $N$ where the two functions agree. Applying induction with $\text{(2)}$, it is immediately evident that $M = N$. $\quad \blacksquare$
Theorem 2: For any $m \in N$, there exist a $\mu \in \mathcal C$ such that $\mu(1) = m$.
Proof
A̶g̶a̶i̶n̶,̶ ̶s̶i̶m̶p̶l̶y̶ ̶a̶p̶p̶l̶y̶ ̶i̶n̶d̶u̶c̶t̶i̶o̶n̶.̶
See this.
$\quad \blacksquare$
So $\mathcal C$ is in bijective correspondence with $N$.
Theorem 3: Let $(N, \sigma)$ and $(N', \sigma')$ be two Peano systems and $\mathcal C$ and $\mathcal C'$ the corresponding semigroups. Then there exist one and only one bijective correspondence $\beta: N \to N'$ satisfying
$\tag 3 \beta(1) = 1'$
$\tag 4 \beta \circ \sigma = \sigma' \circ \beta$
Proof
The function $\beta$ is defined using recursion. Induction is used to show that $\beta$ is injective. Induction is used to show that $\beta$ is surjective. $\quad \blacksquare$
Using the above an argument can be supplied to prove the following.
Theorem 4: Let $(N, \sigma)$ and $(N', \sigma')$ be two Peano systems and $\mathcal C$ and $\mathcal C'$ the corresponding semigroups. Then the mapping $\sigma \mapsto \sigma'$ can be extended to an algebraic isomorphism between $\mathcal C$ and $\mathcal C'$.
We reserve the symbol $\mathbb N$ to denote $\mathcal C$ and use the symbol $+$ to denote the binary operation of composition.
Using the axioms of $ZF$, the existence of Peano systems is no problem.
I couldn't find this technique here or on wikipedia, prompting
Is the theory described above coherent?
If it is it would certainly appeal to students who like to see some 'motion/action' when studying mathematical constructions, say somebody born to be a functional analyst.
Since there has been no feedback except for two upvotes, the theory is sound. So here we knock off the remaining properties of $(\mathbb N,0,1,+)$ that we need to have under our belts. When this is competed, these properties will completely define the natural numbers under addition up to isomorphism.
Note that when convenient, we can always regard $0 \in \mathbb N$ as the identity mapping on $N$ and view the remaining elements in $\mathbb N$ as (proper) injections.
Define the function $\alpha$ on $\mathbb N$ by $k \mapsto k +1$.
Theorem 5: $(\mathbb N, 0, \alpha)$ is a Peano system.
Proof (sketch)
Translating back to $\mathcal C$, the identity fucntion can't have the form $\sigma \circ \tau$, since $\sigma$ can't have a left and right inverse. So $\text{(1)}$ is satisfied. To show $\text{(2)}$, you use induction on $N$ and theorem 1 and theorem 2 to show that there is 'complete coverage'. $\quad \blacksquare$
The prior theorem is stating that $(\mathbb N, 0, \alpha)$ can be viewed as the 'dual Peano system' to $(N, 1, \sigma)$.
Theorem 6: For $z,x,y \in \mathbb N$, if $z + x = z + y$, then $x = y$.
Proof
Injective functions always have a left inverse. $\quad \blacksquare$
Definition: For $m, n \in \mathbb N$, we write $m \le n$ if there exist a $k \in \mathbb N$ such that $m + k = n$.
Theorem 7: The relation $(\mathbb N, \le)$ is a total ordering.
Proof (sketch)
At all steps, when necessary invoke theorem 6 and use algebraic manipulations. Transitivity is straightforward. To show antisymmetry, use the fact that $m + n = 0$ implies that both $m$ and $n$ are equal to $0$. To show the connex property, use induction and split out the logic. $\quad \blacksquare$
Theorem 8: Every nonempty subset of $\mathbb N$ has a least element.
