Suppose f : A → B, g : B → C are both surjections. Prove thatg ◦ f : A → C is a surjection

To prove that the composition of two surjective functions is also surjective, we need to show that for every element 'c' in the set C, there exists an element 'a' in the set A such that (g ◦ f)(a) = c.

Here are the steps:

1. Since g : B → C is surjective, for every 'c' in C, there exists an element 'b' in B such that g(b) = c.

2. Now, since f : A → B is surjective, for this 'b' in B, there exists an element 'a' in A such that f(a) = b.

3. Substituting f(a) = b into g(b) = c, we get g(f(a)) = c, which is the same as (g ◦ f)(a) = c.

Therefore, for every 'c' in C, there exists an 'a' in A such that (g ◦ f)(a) = c, which means that g ◦ f : A → C is surjective. This completes the proof.