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

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If g(f(x)) is injective, then f must also be injective. The proof begins by assuming g(f(x1)) = g(f(x2)) leads to x1 = x2. By showing that if f(x1) = f(x2), then g(f(x1)) = g(f(x2)), the injectivity of g(f) implies x1 must equal x2. This establishes that f is indeed injective. The discussion concludes that the injectivity of the composition g(f) guarantees the injectivity of f.
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.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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