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i have a question

let $E,F$ and $G$ three sets.

$f:E\rightarrow F$

$g:F\rightarrow G$

$h:G\rightarrow E$

prove that if : $f \circ h\circ g$ and $h \circ g\circ f$ are injectives and $g \circ f\circ h$ is surjective. then $f,g,h$ are bijectives.

i have proved that:

$g$ is injective and surjective.

and $g\circ f$ is injective and surjective.

how i can prove that $f$ and $ h$ are bijectives

please help me with this question

Moni145
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ana nadi lwa3r
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1 Answers1

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Since $f\circ h\circ g$ and $h\circ g\circ f$ are injective, $g$ and $f$ are injective. And, since $g\circ f\circ h$ is surjective, $g$ is surjective. Therefore, $g$ is a bijection.

But, since $g$ is a bijection and $f\circ h\circ g$ is injective, $f\circ h$ is injective. Therefore, $h$ is injective. And since $g$ is a bijection and $g\circ f\circ h$ is surjective, $f\circ h$ is surjective. Therefore, $f$ is surjective.

Since $f$ is injective and surjective, it is a bijection. And since $f$ is a bijection and $f\circ h$ is a bijection too, $h$ is a bijection.

Another User
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  • thanks for the help.But i think that your proof is not rigorous. – ana nadi lwa3r Sep 30 '23 at 21:39
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    Here are been used results that are quite known, regarding functions composition btw if @ananadilwa3r you need it you can fill the missing pieces helping you with this example – Turquoise Tilt Sep 30 '23 at 21:47
  • how do you proved that if $g$ is a bijection and $f\circ h\circ g$ is injective then $f\circ h$ is injective – ana nadi lwa3r Sep 30 '23 at 21:50
  • If $f\circ h$ is not injective, then there are distinct elements $y$ and $y'$ in $G$ such that $f(h(y))=f(h(y'))$. Since $g$ is surjective, there are elements $x$ and $x'$ in $F$ such that $g(x)=y$ and that $g(x')=y'$ and, since $y\neq y'$, $x\neq x'$. But then $f(h(g(x)))=f(h(g(x')))$. That's impossible, since $f\circ h\circ g$ is injective and $x\neq x'$. – Another User Sep 30 '23 at 22:05
  • thanks for the help – ana nadi lwa3r Sep 30 '23 at 22:09
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    @ananadilwa3r the key facts being used here are: 1. Compositions of injective maps are injective, and similarly for compositions of surjective maps. 2. the first function in a surjective composition is surjective, 3. the last function in an injective composition is injective. 4. Bijective functions have inverses. All of these would be good exercises to prove. In particular, you know $f\circ h = (f\circ h\circ g)\circ g^{-1}$ (from 4), and you know the right hand side is injective (from 1). – M W Sep 30 '23 at 22:12