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)? 

It might help to give not another answer alongside Xoff's, but to give some motivation for expecting the answer to your question to be 'False'.
Suppose $f\colon \mathbb{N} \to \mathbb{N}$ be bijective and primitive recursive. Then we know $f^{-1}\colon \mathbb{N} \to \mathbb{N}$ is defined. And we can show that it is computable. For here's the obvious method of computation (the one that convinces us that $f^{-1}$ is indeed computable). To calculate $f^{-1}(n)$, compute in turn $f(0), f(1), f(2), \ldots, f(k)$ until we hit $k$ such that $f(k) = n$, then put $f^{-1}(n) = k$. That's a computation, since each of $f(j)$ is finitely computable, and it terminates since $f$ is bijective on the numbers so eventually for some $k$, $f(k) = n$.
But note: 

This computation [and obvious variants, e.g. by interweaving steps of the calculation of the various different $f(j)$] requires an open-ended do until loop, and
  primitive recursive functions can all be computed using only closed for
  $x = 1$ to $j$ loops.

That's not yet a proof. But it tells us why, morally, we can't expect $f^{-1}$ to be primitive recursive in general.   
