Answer :

Define f: N → Z as f(x) = x and g: Z → Z as g(x) = |x|

Now, we can see that

g(-1) = |-1| = 1

g(1) = |1| = 1

⇒ g(-1) = g(1) , but -1 ≠ 1

⇒ g is not injective.

Now, gof: N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|

Let x, y ϵ N such that gof(x) = gof(y).

⇒ |x| = |y|

⇒ x = y

Therefore, gof is injective.

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