Inverse of a primitive recursive bijection Is it true or false that the inverse of a primitive recursive bijection $f: \mathbb{N} \to \mathbb{N}$ is also primitive recursive (pr)? 

No this is not true!
Let's $A$ the Ackermann-Peter function (defined by $A(x)=A(x,x)$ with this definition). Note that $A$ is fast growing, recursive but not primitive recursive. 
But $A^{-1}$ defined by $A^{-1}(x)=1+\max\{y\;|\;A(y)\le x\}$ ($0$ if the set is empty) is recursive primitive (and very slow growing).
Now consider the sets $B=\{A(x) \;|\; x \in \mathbb N\}$ and $\overline{B}=\mathbb N\setminus B $ (both infinite).
And the sets $Odd=\{2x+1 \;|\; x \in \mathbb N\}$ and $\overline{Odd}=\mathbb N\setminus E $ (both infinite).
Then the bijection $\mathcal B$ that matchs the $k^{th}$ element of $B$ to the $k^{th}$ element of $Odd$ and the $k^{th}$ element of $\overline{B}$ to the $k^{th}$ element of $\overline{Odd}$ is recursive primitive (because given $x$, you need to compute $y=A^{-1}(x)$ and at the same time verify that $A(y-1)=x$ (both can be done with primitive functions). if $A(y-1)=x$ then $\mathcal B(x)=2y-1$ else $\mathcal B(x)=2*(x-y)$
But $\mathcal B^{-1}$ is not primitive recursive as $\mathcal B^{-1}(2x+1)=A(x)$
