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Given $$(f◦g)(x)=x$$ (from $\Bbb R$ to $\Bbb R$ for any $x$ in $\Bbb R$)
And $g(x)$ is onto!

does it mean that also $$(g◦f)(x)=x$$

It seems like it does but how can I prove it?

Khashayar Baghizadeh
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The One
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  • http://en.wikipedia.org/wiki/Surjective_function#Surjections_as_right_invertible_functions –  Dec 14 '14 at 18:50
  • From your other question you know that $g$ is injective; add sujectivity to that and you know that there exist in inverse function $h$ such that $g(h(x))=h(g(x))=x$, next you show that $h=f$ and from this it follows that $g(f(x))=x$. – Myself Dec 14 '14 at 18:52
  • @Myself but I think that g is not injective at all – The One Dec 14 '14 at 18:53
  • Either because you have learned it in some class, or because you are able to show it: for each $x$ there is a unique $y$ such that $y=g(x)$. So the definition $h(y)=x$ makes sense. Then show, using this definition, that $g(h(x))=h(g(x))=x$. – Myself Dec 14 '14 at 18:57
  • @Myself I dont have a teacher I need to learn everything alone :( so I'll take it as axiom – The One Dec 14 '14 at 19:02

1 Answers1

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It seems the following.

Since the map $g$ is onto, $x=g(y)$ for some $y\in R$. Then $$(g\circ f)(x)= (g\circ f)(g(y))=g\circ f\circ g(y)=g(y)=x.$$

Alex Ravsky
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