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

We first show that g is not injective.

It can be observed that:

g(– 1)=| – 1| = 1

g(1) =|1| = 1

Therefore, g(– 1) = g(1), but —1 ≠ 1.

Therefore, g is not injective.

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

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

⇒ |x|=|y|

Since x and y ∈N both are positive.

∴ |x|=|y| ⇒ x=y

Hence, gof is injective