Let f be a function from a set A to a set B, g a function from B to C, and h a…

Correct statement: The statement "h is onto => g is onto" is always true.

Proof:

• Assume h = g∘f is onto. For any element c in C there exists an a in A with h(a) = c, i.e. g(f(a)) = c.

• Let b = f(a). Then b belongs to B and g(b) = c. Since c was an arbitrary element of C, this shows that for every c in C there exists some b in B with g(b) = c, so g is onto.

Why the other statements fail (short counterexamples):

• The statement "g is onto => h is onto" can fail. Example: A = {a}, B = {b1, b2}, C = {c1, c2}; let f(a) = b1, and let g(b1) = c1, g(b2) = c2. Then g is onto but h maps A only to c1, so h is not onto.

• The statement "h is onto => f is onto" can fail. Example: A = {a}, B = {b1, b2}, C = {c1}; set f(a) = b1 and g(b1) = g(b2) = c1. Then h is onto C but f does not hit b2, so f is not onto.

• The statement "h is onto => f and g are onto" is false because f need not be onto; the previous counterexample shows h onto and g onto while f is not onto.

Takeaway: If the composition h = g∘f is surjective onto C, then g must be surjective onto C, but f need not be surjective onto B.