\(π(. )\) is a function from \(A\) to \(π΅, π(. )\) is a function from \(B\)β¦
2025
\(π(. )\)Β is a function fromΒ \(A\) toΒ \(π΅, π(. )\) is a function fromΒ \(B\) to \(C\), and their composition defined asΒ \(π(π(. ))\) is a mapping fromΒ \(A\) to \(C\). IfΒ \(π(. )\) andΒ \(π(π(. ))\) are onto (surjective) functions, which ONE of the following is TRUE about the function \(π(. )\)?
Answer: D. \(π(. )\) is not required to be a one-to-one or onto function. β Correct conclusion: g is not required to be a one-to-one or onto function. Explanation: If fβg is onto C, then for every c in C there exists some a in A withβ¦
- A.
\(π(. )\)Β must be an onto (surjective) function. - B.
\(π(. )\)Β must be a one-to-one (injective) function. - C.
\(π(. )\)Β must be a bijective function, that is, both one-to-one and onto. - D.
\(π(. )\)Β is not required to be a one-to-one or onto function.
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Show answer & explanation
Correct answer: D
Correct conclusion: g is not required to be a one-to-one or onto function.
Explanation:
If fβg 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 fβg(a) = c, so fβg 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 fβg maps both a1 and a2 to c, so fβg 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.