Let g be a function from the set A to the set B and let f be

Let g be a function from the set A to the set B and let f be a function from the set B to the set C. Let h be composition of f and g, i.e., h = f g.

1. Prove that if h is one-to-one, then g must be one-to-one. Give an example to show that f is not necessarily one-to-one.

2. Prove that if h is onto, then f is onto. Give an example to show that g is not necessarily onto.

Solution

(1). Suppose that h = fog is injective (one-one); we show that g is injective.

let x₁, x₂ belongs to A and suppose that g(x₁) = g(x₂). Then

(fog)(x₁) = f(g(x₁)) = f(g(x₂)) = (fog)(x₂).

But since fog is injective, this implies that x₁ = x₂. Therefore f is injective.

Next, we prove (2).

Suppose that h = fog is surjective. Let z belongs to C.

Then since h = fog is surjective,

there exists x belongs to A such that (fog)(x) = f(g(x)) = z. Therefore if we let

y = g(x) belongs to B,

then f(y) = z. Thus f is surjective.

The same example works for both. Let A = {1}, B = {1, 2} C = {1} and

g : A to B by g(1) = 1 and f : B to C by f(1) = f(2) = 1.

Then fog : A to C is defined by (fog)(1) = 1.

This map is a bijection from A = {1} to C = {1}, so is injective and surjective. However, f is not injective, since f(1) = f(2) = 1, and g is not surjective,

since 2 does not belongs to g(A) = {1}.