Poly-time computability of inversion of poly-time real functions At pp. 7-8 of Ker-I Ko's Computational Complexity of Real Functions (1991), the following is stated for one dimensional cases:


  Let $INV_1$ be the operator that maps a one-to-one function $f:[0,1]\rightarrow [0,1]$ to its inverse function $f^{-1}$. Then $INV_1(f)$ is polynomial-time computable for all polynomial-time computable, one-to-one real functions $f$ on $[0,1]$.


And for two dimensional cases:


  Let $INV_2$ be the operator that maps a one-to-one function $f:[0,1]^2\rightarrow [0,1]^2$ to its inverse function $f^{-1}$. Then $P=NP$ implies that for all polynomial computable, one-to-one real functions $f$ on $[0,1]^2$, $INV_2(f)$ is polynomial-time computable, and this in turn implies $P=UP$.


How is the one-dimensional case derived? And how can the two-dimensional case be extended to n-dimensions?
In 1-dimensional case, the function has to be strictly-monotone. 
Given $y_n$ do a binary search to find $x_n$.

If $\mathsf{P} = \mathsf{NP}$, 
given $y_n$ we can guess $x_n$ and then verify that $f(x_n) = y_n$.
The dimension doesn't matter for this part.

For the other part, that is, if $INV_d$ is polytime then $\mathsf{P} = \mathsf{UP}$,
we can reduce it to $INV_2$ by noting that every 2-dimensional function 
can be extended to a $d$-dimensional one by letting other coordinates map as identity.
So for $d'&lt;d$,
$INV_d$ being polytime computable implies that $INV_{d'}$ is also polytime.
Therefore $INV_d$ being polytime implies that $INV_2$ is polytime and therefore $\mathsf{P}=\mathsf{UP}$.