I have a little problem. I'm probably being just so careless..... Here k-varieties are all integral separated k-schemes of finite type over k, where k is a field.


Suppose $X, Y$ are $k$-varieties, and let $f :X \to Y$ be a morphism of $k$-varieties that is one to one and onto. Then, when can we say this $f$ is an isomorphism of $k$-varieties?


If this is too vague, let me add that the case I would like to see is when each fibre of $f$ (which is a singleton) is reduced. Under this assumption, would this give an isomorphism?



When two k-varieties with the same underlying topological spaces isomorphic?
The condition you are looking for is **seminormality**. A variety (or a reduced scheme) $Y$ is *seminormal* if any proper bijective morphism $f:X\to Y$, with $X$ reduced, *inducing isomorphisms on residue fields* $k(y)=k(x)$ for points $x\in X$, $y=f(x)\in Y$, is an isomorphism. A basic fact is that any variety has a unique seminormalization. 


A related notion which differs only in positive characteristic is **weak normality** for which $k(y)\to k(x)$ is required to be purely inseparable and an isomorphism for each *generic point* $x\in X$.


One basic reference for this notion is the appendix to Chapter 1 of Koll'ar's "Rational curves on algebraic varieties", where you will find many standard facts and examples such as: normal implies seminormal; in dim 1 seminormal means analytically isomorphic to the $n$ axes in $A^n$; irreducible components of seminormal schemes need not be normal, etc. You will also find references to many papers where this notion was comprehensively investigated.


For clarity, let me add the standard fact: $f$ is proper and bijective $\iff$ it is *finite* and bijective (as opposed to quasifinite = finite fibers, which of course follows from bijective).


