\(𝑔(. )\) is a function from \(A\) to \(𝐵, 𝑓(. )\) is a function from \(B\)…

Correct conclusion: g is not required to be a one-to-one or onto function.

Explanation:

If f∘g is onto C, then for every c in C there exists some a in A with f(g(a)) = c. This means f maps the image of g (a subset of B) onto C, i.e. f(image(g)) = C. It does not force image(g) to equal all of B, nor does it force g to be injective.

• Counterexample showing g need not be onto: Let C = {c}, B = {b1, b2}, A = {a}. Define f(b1) = c and f(b2) = c (so f is onto C). Define g(a) = b1. Then f∘g(a) = c, so f∘g is onto C, but g does not hit b2 and so is not onto B.

• Counterexample showing g need not be injective: Let B = {b}, C = {c}, A = {a1, a2}. Define f(b) = c (so f is onto). Define g(a1) = g(a2) = b. Then f∘g maps both a1 and a2 to c, so f∘g is onto, but g is not injective.

Key point: what must hold is that the image of g is large enough that f(image(g)) = C; g itself need not be onto B or one-to-one.