Let `A={-1,\ 0,\ 1}` and `f={(x ,\ x^2): x in A}` . Show that `f: A->A` is neither one-one nor onto.

Explanation

Injectivity:

Let `x = 1`

Therefore `f(1) = 1^2 = 1` and `f(-1) = (-1)2 = 1`

` ⇒ 1` and `-1` have the same images.

Therefore, `f` is not one-one.

Surjectivity:

Co-domain of `f = {-1, 0, 1}`

` f(1) = 1^2 = 1, f(-1) = (-1)^2 = 1` and `f(0) = 0`
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