For nonempty sets A, B, C, let f: (A -->B) and g: (B-->C) be functions, Prove by

Question

For nonempty sets A, B, C, let f: (A -->B) and g: (B-->C) be functions,

Prove by a direct proof, contrapositive, and contradion of:

If (g0f) is one to one (injective), then f is one to one (injecive).

Explanation / Answer

1. suppose f(x) = f(y) -->g(f(x) = g(f(y) but gof is 1:2 --> x= y 2. suppose the composition of g and f is not 1:1 then exist x,y not equal s.t. g(f(x)) = g(f(y)) -->f(x) = f(y) so f is not 1:1 3. Suppose f is not 1:1 then there exists x,y not equal st f(x)=f(y) --> g(f(x)) = g(f(y)) but x /= y so g o f is not 1:1 contradiction

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.