The induction axiom is independent of the other Peano axioms

P1 holds: $\sigma$ is the successor within each copy; $0$ is the $\mathbb{N}$-copy zero, which is not the successor of any element (nothing in $N_{\mathbb{N}}$ maps to it, and $\sigma$ maps $N_{\mathbb{Z}}$ into $N_{\mathbb{Z}}$), so $\sigma (x)\mathrm{\ne }0$ for all $x$.

P2 holds: $\sigma$ is injective on $N_{\mathbb{N}}$ and on $N_{\mathbb{Z}}$ separately, and $\sigma$ maps each copy into itself, so $\sigma$ is injective on $N$.

P3 fails: let $S=N_{\mathbb{N}}$, the $\mathbb{N}$-copy; then $0\in S$ and $\sigma$ maps $S$ into $S$ ($\sigma (N_{\mathbb{N}})\subseteq N_{\mathbb{N}}$), so $S$ contains $0$ and is closed under $\sigma$; the $\mathbb{N}$-copy is the standard model [L2].

But $S=N_{\mathbb{N}}\mathrm{\ne }N$, since the $\mathbb{Z}$-copy $N_{\mathbb{Z}}$ is disjoint from $S$ and nonempty.

Thus $(N,0,\sigma )$ satisfies P1 and P2 but not P3, refuting the claim: induction is independent of P1 and P2 and cannot be dropped.