Successor Mapping on Natural Numbers is not Surjection

Theorem
Let $f: \N \to \N$ be the successor mapping on the natural numbers $\N$:


 * $\forall n \in \N: \map f n = n + 1$

Then $f$ is not a surjection.

Proof
There exists no $n \in \N$ such that $n + 1 = 0$.

Thus $\map f 0$ has no preimage.

The result follows by definition of surjection.