Discussion
\(g(.)\) is a function from \(A\) to \(B\), \(f(.)\) is a function from \(B\) to \(C\), and their composition defined as \(f(g(.))\) is a mapping from \(A\) to \(C\).
If \(f(.)\) and \(f(g(.))\) are onto (surjective) functions, which ONE of the following is TRUE about the function \(g(.)\)?
Answer: D
The correct statement is \(g(.)\) is not required to be a one-to-one or onto function.
That \(f \circ g\) is onto forces f to be onto, but places no requirement on g. A counterexample shows both properties can fail. Let \(A = \{1, 2\}\), \(B = \{x, y, z\}\), \(C = \{c\}\). Take \(g(1) = g(2) = x\), and let f send everything to c.
Then f and \(f(g(.))\) are onto C, yet g is not one-to-one (1 and 2 both go to x) and not onto (nothing maps to y or z). So g need not be injective, surjective or bijective, and options A, B and C, which say it must be, are false.
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