Prove that if g(f(x)) is injective then f is injective

[cbarker1]

Dear Everybody,
Question:
"Prove that if g(f(x)) is injective then f is injective"
Work:
Proof: Suppose g(f(x)) is injective. Then g(f(x1))=g(f(x2)) for some x1,x2 belongs to C implies that x1=x2. Let y1 and y2 belongs to C. Since g is a function, then y1=y2 implies that g(y1)=g(y2). Suppose that f(x1)=f(x2). Then g(f(x1))=g(f(x2)). Therefore f is injective. QED

 

[Euge]

You haven't proved $f$ is injective. To fix it, suppose $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Injectivity of $g\circ f$ implies $x_1 = x_2$. Thus $f$ is injective.

 

[math771]

.
Great proof! Your explanation is clear and concise. One small suggestion I have is to define what C represents in the proof, just for clarity. Other than that, well done! Keep up the good work.