I have two rather distinct questions on rational functions and rations maps:
1) Let $X$ be an affine variety (irreducible, if necessarily) and $U \subset X$ -- a nonempty open set. $f \in K(X)$ -rational function, with $dom{f}=U$. Is it possible to find $g \in K[X]$ -- regular function on $X$ such that $f$ and $g$ coincides on $U$?
2) Let $\phi: X \rightarrow Y$ be dominant morphism of X, Y -- irreducible affine varieties. $\phi$ is injective on some open set $X_0$. It is said then that for existence of rational map $\rho: Y \rightarrow X$ such that $\rho\phi=1_{X}$ (everywhere where $\rho$ is defined) it is sufficient to have an inclusion $K[X] \hookrightarrow K(Y)$. So, my question is natural: why is it sufficient?
As I see, such inclusion $K[X] \hookrightarrow K(Y)$ (denote it by $\rho^*$) defines dominant rational map $\rho: Y \rightarrow X$. Since $\rho$ is defined on some open subset of $Y$, $\phi(X_0)$ contains principal open set of $Y$, and the fact that any two open subsets of irr. varieties intersect -- the composition $\rho\phi$ is well defined on some open subset $U$ of $X$ ($U \subset X_0$). The question is how to make $\rho\phi$ be identity there (why will it automatically be identity)?
Many thanks for any explanations!
UPD: The answer to the question 1) is affirmative iff $f$ is regular.