are fibres of a flat bijective map reduced? Let $f: X \to Y$ be a flat map of algebraic varieties or of complex analytic spaces which is bijective on closed points (or just bijective in the secnond case). Suppose  both $X$ and $Y$ are reduced. Is it true that $f$ has reduced fibres?
If it is true, I would be most grateful for a reference. 
 A: Let $f: X\to Y$ be a bijective flat morphism of reduced algebraic varieties over $\mathbb C$ (or any algebraically closed field $k$ of characteristic $0$), then $f$ is an isomorphism. 
First $f$ is quasi-finite, hence finite and étale (because characteristic $0$) above some dense open subset $V$ of $Y$. As we work over an algebraically closed field, $f^{-1}(V)\to V$ is then an isomorphism. 
Let $x\in X$ and $y=f(x)$. Then $O_{Y,y}\to O_{X,x}$ is flat, hence faithfully flat, therefore injective. This implies that the quotient $O_{X,x}/O_{Y,y}$ is flat over $O_{Y,y}$. But the total rings of fractions of $O_{Y,y}$ and $O_{X,x}$ coincide because $X\to Y$ is birational by the above. So $O_{X,x}/O_{Y,y}$ is of torsion over $O_{Y,y}$, hence equal to $0$. So $f$ is an open immersion. But $f$ is surjective, it is an isomorphism. 
The proof should work for reduced complex analytic spaces. 
A: If $D=Spec(\mathbb C[T]/T^2)$ is the double point and $P$ is the simple point, the morphism of analytic spaces $D\to P$ has reduced base, is flat and  bijective but has non reduced fiber.
