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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.2012-12-15

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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!2012-12-15
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    Yes I didnt see that2012-12-15
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    Note that this works even if $g$ is not assumed to be injective.2012-12-15
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    @pie you can [accept](http://meta.math.stackexchange.com/a/3287/8271) this answer.2012-12-15
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    On another glace, I don't understands how you reached the conclusion a=b... can you explain please? Because if f(x)=x^2, then f(a)=f(b) (for instance a=2 and b=-2) and g(f(a))=g(f(b)) but b=/=a. Maybe I missed something?2012-12-15
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    If $f(x)=x^2$, then $g\circ f$ will not be injective (on $\mathbb{R}$).2012-12-15
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    sorry then I don't understand why since g∘f is injective, a=b.2012-12-15