a. Let f:A-->B and g:B-->C be functions. Prove that if g of f is injective, then

Question

a. Let f:A-->B and g:B-->C be functions. Prove that if g of f is injective, then f is injective.

b. Is the converse true? If so, prove it. If not, give a counterexample.

Explanation / Answer

proof by contradiction let f not be injective ==> we have a,b such that f(a) = f(b) = x ==> g(x) = g(f(a) = g(f(b) ) ==> gof is not injective but given that gof is injective this is a contradiction hence f ios injective part b no converse is not true for example let f(x) = x and g(x) = x^2 clearly f is injective but gof is noit because gof(1) =gof(-1) hence converse is not true always

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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