Classify `f: Z->Z` given by `f(x)=x^2` as injection, surjection or bijection.

Explanation

Injection condition:

Let `x` and `y` be any two elements in domain `(Z)`, such that `f(x) = f(y)`.

`f(x) = f(y)`

`x^2 = y^2=> x = pmy`

Therefore, f is not an injection.

Surjection test:

Let `y` be any element in the co-domain `(Z)`, such that `f(x) = y` for some element `x` in `Z` (domain).

`f(x) = y=> x^2 = y`
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