Check the injectivity and surjectivity of the following functions: f: Z → Z given by f(x) = x3
f: Z → Z is given by,
f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.