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Conjecture: If $g(x)$ is injective, and $g(f(x))$ is injective, then $f(x)$ is injective

How can I prove that conjecture formally?

Thanks!

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I think you should specify domains and codomains of f and g. –  Moritzplatz Dec 15 '12 at 15:03

1 Answer 1

up vote 2 down vote accepted

Let $f(a)=f(b)$. Hence $g(f(a))=g(f(b))$. Since $gf$ is injective. Therefore $a=b$

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Very nice Amr. Thanks! –  pie Dec 15 '12 at 15:02
    
Yes I didnt see that –  Amr Dec 15 '12 at 15:04
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Note that this works even if $g$ is not assumed to be injective. –  Santiago Canez Dec 15 '12 at 15:07
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@pie you can accept this answer. –  leo Dec 15 '12 at 15:14
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If $f(x)=x^2$, then $g\circ f$ will not be injective (on $\mathbb{R}$). –  Benji Dec 15 '12 at 15:20

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