Let a function f: ℕ → ℕ be defined by f(n) = [ cases 2n, & n = 2,...

For one value of $n$ we will get only one corresponding value of $f\left(n\right)$, so $f\left(n\right)$ is one-one

Now, for $n = 2, 4, 6 ⋯ ; f\left(n\right) = 4, 8, 12 . . .$

for $n = 3, 7, 11 ⋯ ; f\left(n\right) = 2, 6, 10 . . .$

for $n = 1, 5, 9 ⋯ ; f\left(n\right) = 1, 3, 5, 7 . . .$

So range of $f\left(n\right)$ is $N$

Hence $f\left(n\right)$ is onto