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Question
Let A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.
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Solution
A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}
Given, f(x) = x2
Injectivity :
f(1) = 12=1 and
f(-1)=(-1)2=1
⇒ 1 and -1 have the same images.
So, f is not one-one.
Surjectivity :
Co-domain of f = {-1, 0, 1}
f(1) = 12 = 1,
f(-1) = (-1)2 = 1 and
f(0) = 0
⇒ Range of f = {0, 1}
So, both are not same.
Hence, f is not onto
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