Let A→B and B→C denote two functions. If the function
Let f: A→B and g: B→C denote two functions. If the function gof: A→C is a surjective then
- A.
g is into but f is onto
- B.
Both must be into
- C.
g is onto but f need not to be onto
- D.
f is onto but g need not to be onto
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Show answer & explanation
Correct answer: C
Key fact: If g∘f is surjective, then g must be surjective.
Reason: For any element c in C, since g∘f is surjective there exists an a in A with g(f(a)) = c. Let b = f(a) in B. Then g(b) = c, so every element of C has a preimage in B. Hence g is surjective.
About f: f need not be surjective. A counterexample makes this clear.
Take A = {1}, B = {1,2}, C = {x}.
Define f by f(1) = 1 (so f is not onto B because 2 has no preimage). Define g by g(1) = x and g(2) = x (so g is onto C).
Then for the only a = 1 we have g(f(1)) = g(1) = x, so g∘f is onto C even though f is not onto B.
Conclusion: The correct statement is that the second function is onto, but the first function need not be onto.