elementary set theory - Showing a function $H:mathbb R^{mathbb R}to mathbb R^{mathbb R}$ is onto and finding its inverse

Let $H:\mathbb R^{\mathbb R}\to \mathbb R^{\mathbb R} \\ H(f)=\begin {cases}f^{-1} & \text {$f$ is a bijection} \\ f & else \end {cases}$

Prove that $H$ is onto (I already showed it's 1-1) and find $H^{-1}$ and prove it.

Onto: Let $f\in \mathbb R^{\mathbb R}$.

Suppose $f$ is a bijection, then there's a single $f^{-1}$ and from the defintion of $H$ we'll get $H(f^{-1})=f$.

Suppose $f$ isn't a bijection, then from $H:$ $H(f)=f$ either way for every $f$ there's a source, therefore, $H$ is onto.

I think it's inverse is simply: $H(f)=\begin {cases}f^{-1} & else \\ f & \text {$f$ is a bijection} \end {cases}$

But I can't seem to be able to compose $H\circ H^{-1}$ and vice versa correctly...

Answer

By definition $\;H\circ H(f)=H(H(f))\;$:

$$H(H(f))=\begin{cases}H(f^{-1})&,\;\;f\;\;\text{is a bijection}=\left(f^{-1}\right)^{-1}=&f&,\;f\;\text{is a bijection}\\{}\\
H(f)&,\;\;f\;\;\text{else}=& f&,\;f\;\text{else}\end{cases}$$