Let be function from to and be function from to

Concept: for functions f : A → B and g : B → C, the only well-formed composition here is g∘f : A → C, given by (g∘f)(x) = g(f(x)) (f∘g is not well-formed, since g's codomain C need not equal f's domain A). Four composition-transfer facts govern such problems: (i) if g∘f is onto, g is onto; (ii) if f and g are both onto, g∘f is onto; (iii) if f and g are both one-to-one, g∘f is one-to-one; (iv) if f and g∘f are only one-to-one (without f being onto B), g need not be one-to-one.

Applying these four facts to the given statements: three of the options restate facts (i)-(iii) exactly, which always hold. The remaining option claims that f and g∘f being one-to-one forces g to be one-to-one — this is fact (iv)'s failure case, since that guarantee needs the additional condition that f is onto B, which is not assumed.

Counterexample (cross-check that this property genuinely fails):

• Take sets A = {1}, B = {1,2}, C = {0}. Define f(1) = 1 (so f is one-to-one) and define g(1) = 0, g(2) = 0 (so g is not one-to-one).

• Then g∘f maps 1 to 0 and is one-to-one (trivially, since domain has one element). Thus f and g∘f are one-to-one while g is not, so the property is false.

Cross-check — proofs that the other three properties hold:

• If the composition g∘f is onto, then g is onto. Proof: for any c in C choose a in A with g(f(a)) = c; letting b = f(a) ∈ B gives g(b) = c.

• If f and g are both onto, then g∘f is onto. Proof: for any c in C choose b in B with g(b) = c and then a in A with f(a) = b; then g(f(a)) = c.

• If f and g are both one-to-one, then g∘f is one-to-one. Proof: if g(f(x1)) = g(f(x2)) then injectivity of g gives f(x1)=f(x2), and injectivity of f gives x1=x2.

Result: the incorrect property is the one asserting that f and g∘f one-to-one alone force g to be one-to-one.