# Proving bijectivity of a linear mapping

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.

• If the answers below worked for you personally, please accept the one you are most satisfied by clicking on the check-mark on the left of the answer. Commented Jan 18 at 15:00

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.