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Question
Check the injectivity and surjectivity of the following function:
f: Z → Z given by f(x) = x3
Solution
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.
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