Suppose f : A → B and g : B → C are both surjective functions.Prove that g ◦ f : A → C is surjective
Question
Suppose f : A → B and g : B → C are both surjective functions.
Prove that g ◦ f : A → C is surjective.
Solution
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:
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Since g : B → C is surjective, for every 'c' in C, there exists an element 'b' in B such that g(b) = c.
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Now, since f : A → B is surjective, for this 'b' in B, there exists an element 'a' in A such that f(a) = b.
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Substituting f(a) = b into g(b) = c, we get g(f(a)) = c, which is the same as (g ◦ f)(a) = c.
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Therefore, for every 'c' in C, we have found an 'a' in A such that (g ◦ f)(a) = c.
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Hence, g ◦ f : A → C is surjective.
This completes the proof.
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