Check the injectivity and surjectivity of the following functions:
f : Z → Z given by f (x) = x3
It is given that f : Z → Z given by f (x) = x3
We can see that for x, y ϵ N,
f(x) = f(y)
⇒ x3 = y3
⇒ x = y
⇒ f is injective.
Now, let 2 ϵ Z. But, we can see that there does not exists any x in Z such that
f(x) = x3 = 2
⇒ f is not surjective.
Therefore, function f is injective but not surjective.