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

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SUMMARY

The discussion centers on proving that if the composition of functions g(f(x)) is injective, then the function f must also be injective. The proof demonstrates that if g(f(x1)) equals g(f(x2)), it leads to the conclusion that x1 equals x2, thereby establishing the injectivity of f. The argument is structured logically, confirming that the injectivity of the composite function necessitates the injectivity of the inner function f. This conclusion is reached through a clear application of definitions and properties of injective functions.

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cbarker1
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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
 
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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.
 

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