Let $f : \mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a function.

$Rf$ : $\mathbb{R}^{\mathbb{R}}$ $\rightarrow$ $\mathbb{R}^{\mathbb{R}}$ is definded by $( g \mapsto g \circ f )$.

I need to show that $Rf$ is bijective exactly when $f$: $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ is.

I already proved that $Rf$ is linear. My approach is to show injectivity and surjectivity separately with the definition: $Rf$ is injective when $\ker Rf = \{0\}$ and $Rf$ is surjective when $\operatorname{im} f$ = $\mathbb{R}$, but I don't know whether I can prove it that way because I have no further information on $g$.

Help would be appreciated.

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2 Answers 2


I find it easier to follow the approach below.

Since $f$ is bijective, then there is a function $f^{-1}:\Bbb R\to\Bbb R$ such that $$f\circ f^{-1}=\operatorname{Id_\Bbb R} \ \ \text{and} \ \ f^{-1}\circ f=\operatorname{Id_\Bbb R}.$$ Now, let $R^{-1}_f:\Bbb R^\Bbb R\to\Bbb R^\Bbb R$ be a function defined by $$R_f^{-1}(g)=g\circ f^{-1}.$$ We can show that \begin{align} R_f\circ R_f^{-1}(g)&=R_f(R_f^{-1}(g))=R_f(g\circ f^{-1})\\ &=(g\circ f^{-1})\circ f=g\circ (f^{-1}\circ f)\\ &=g\circ\operatorname{Id}_{\Bbb R}=g. \end{align} Thus, $R_f\circ R_f^{-1}=\operatorname{Id}_{\Bbb R^\Bbb R}$. Similarly, we can also prove that $R_f^{-1}\circ R_f=\operatorname{Id}_{\Bbb R^\Bbb R}$.

Since $R_f\circ R_f^{-1}=\operatorname{Id}_{\Bbb R^\Bbb R}$ and $R_f^{-1}\circ R_f=\operatorname{Id}_{\Bbb R^\Bbb R}$, we can conclude that $R_f$ is bijective.


Assume that $f$ is bijective. You want to prove that $Rf$ is bijective.

  • Surjectivity
    Given $h\in\mathbb{R}^\mathbb{R}$, you need to find $g\in\mathbb{R}^\mathbb{R}$ such that $g\circ f=h$. Just take $g=h\circ f^{-1}$, which is possible since $f$ is bijective.
  • Injectivity
    If $g\circ f=g'\circ f$, since $f$ is bijective you can compose on the right with $f^{-1}$ and you get $g=g\circ f\circ f^{-1}=g'\circ f \circ f^{-1}=g'$, which proves injectivity.

On the converse, if $Rf$ is bijective you have to prove that so is $f$.
Using the srujectivity of $Rf$, you can find $g\in\mathbb{R}^\mathbb{R}$ such that $g\circ f=1_{\mathbb{R}}$.
But $Rf(f\circ g)=f\circ g \circ f=f\circ 1_{\mathbb{R}}=f=Rf(1_{\mathbb{R}})$. Since $Rf$ is injective you have $f\circ g=1_{\mathbb{R}}$.
Thus $g\circ f=1_{\mathbb{R}}=f\circ g$, which means that $f$ is bijective.


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