Injectivity of the successor is independent of the other Peano axioms

P1 holds: the values of $\sigma$ are $\sigma(0) = 1$, $\sigma(1) = 2$, $\sigma(2) = 2$, none of which is $0$, so $\sigma(x) \neq 0$ for all $x$.

P3 holds: if $S \subseteq N$ with $0 \in S$ and $S$ closed under $\sigma$, then $0 \in S$ forces $1 = \sigma(0) \in S$, then $2 = \sigma(1) \in S$, so $S = \{0, 1, 2\} = N$; thus induction holds.

P2 fails: $\sigma(1) = 2 = \sigma(2)$ but $1 \neq 2$, so $\sigma$ is not injective.

So $(N, 0, \sigma)$ satisfies P1 and P3 but not P2, refuting the claim; note the failure permits the pathology $\sigma(2) = 2$ (a number that is its own successor) and a finite model in which distinct numerals collapse.