Answer:
Check.
Step-by-step explanation:
a. Let [tex]f: \mathbb{Z} \rightarrow O[/tex] such that f(n) = 2n.
Injective:
Let n, m [tex]\in \mathbb{Z}[/tex] such that f(n)=f(m). Then,
2n = 2m and dividing both sides by 2 we obtain that n=m. Then f is injective.
Surjective:
Let n [tex]\in O[/tex], so n = 2k for some integer k. Then, f(k) = n and therefore f is surjective.
b. Let [tex]g: O \rightarrow \mathbb{Z}[/tex] such that g(n) = n/2. Then,
g(f(n)) = g(2n) = 2n/2 = n, and for m [tex]\in O[/tex]
f(g(m)) = f(m/2) = 2*m/2 = m.
Therefore, g is the inverse of f.