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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.

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    My idea was purely intuitive off of the following thoughts: If$f$is unramified, then you do have such an iso. Geometrically, ramification happens when "fibers come together". Your conditions force the map to be of relative dimension 0, and moreover all closed fibers to be singletons, so with that picture in mind fibers can't come together. Thus it seems to me the map is unramified. This could be a situation where my intuition is leading me astray.2012-05-13

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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.

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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.

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    @DimaSustretov, ah sorry. Consider the Frobenius morphism $x\mapsto x^p$ on the affine line over a characteristic p>0 field.2012-05-13