Let $f: X \to Y$ be a regular birational morphism of projective varieties over $k= \mathbb{C}$, $q \in Y$ a point. Claim: If the fiber $f^{-1}(q)$ is disconnected, then $q$ is a singular point of $Y$ (that means Jacobi matrix $(\frac{\partial g_i(q)}{\partial x_j})_{i,j}$ has not maximal rank with $g_1,..., g_d \in \mathbb{C}[x_1,...,x_n]$ cutting out locally $Y \cap \mathbb{C}^n \subset \mathbb{C}^n \subset \mathbb{PC}^n$). In Harris' book Algebraic Geometry, Proposition 16.8 page 204 is clamed that it is immediate over $\mathbb{C}$, follows by Lefschetz principle over any field of cahracteristic $0$, but tricky algebraically.

Why is immediately clear from complex analytic point of view that if the fiber is disconnected $f^{-1}(q)$, then $q$ is singular?

Affine version of the question: Let $X=V(h_1,..., h_k) \subset \mathbb{C}^m, Y=V(g_1,..., g_d) \subset \mathbb{C}^n$ affine complex varieties and $f: X \to Y, x \mapsto (f_1(x),..., f_n(x))$ polynomial map which is bijective on open dense subvarieties $U \subset X, V \subset Y$. By construction this map is continuous with respect Zariski topology and differentiable with respect classical analystic topology.

It seems plausible to expect that the fibers are even analytically connected over smooth points, but I haven't found a right argument.

Edit: Thanks to red_trumpet's counterexample one sees the assumption that $f$ is projective cannot be discarded and the "affine version" suggesed above is plainly wrong.

  • $\begingroup$ I don't think this is true if you drop the projectivity assumption. For example if you start with $Y = \mathbb P^2$, blow up a point $P$ to obtain $X_1 \to Y$. Then blow up a point in the exceptional curve to obtain $X_2 \to Y$. Again blow up a point in the new exceptional curve to obtain $X_3 \to Y$. By construction, the fiber over $P$ consists of three $\mathbb P^1$'s forming a "chain". If you take out the $\mathbb P^1$ in the middle, you obtain a birational morphism $X \to Y$ which has a disconnected fiber over $P$, but $Y$ is smooth. However, $X$ is not projective anymore. $\endgroup$ Feb 21 at 11:26
  • $\begingroup$ @red_trumpet: why such $X$ would not be projective anymore? $\endgroup$
    – user267839
    Feb 23 at 21:24
  • 1
    $\begingroup$ Because it is open in something projective, but not the full thing. As a complex manifold (with the ordinary topology) it will not be compact. Formally, you can choose a curve $C$ in $X_3$ which intersects the middle $\mathbb P^1$ in a point $p$. Then you have a map $C\setminus p \to X$, which cannot be completed to $p$. This contradicts the valuative criterion for properness. $\endgroup$ Feb 23 at 22:09
  • $\begingroup$ @red_trumpet: Thank you for this example, I will modify the question! $\endgroup$
    – user267839
    Feb 23 at 22:34
  • $\begingroup$ If we retain the projectivity of $X,Y$ (and so for $f$ too), there is a deep result that the fibers are connected even if $Y$ is not smooth, but normal. But that's highly non trivial. Do you maybe know if when we assume that $Y$ is smooth, that it becomes more "elementary" or "geometrically" to understand why the fibers are connected? $\endgroup$
    – user267839
    Feb 23 at 22:38

1 Answer 1



Let $f:X \to Y$ be a regular, birational morphism of projective varieties over $\CC$. Further let $q \in Y$ be a regular point. Then $f^{-1}(q)$ is connected.

With the weakening "$q$ is a normal point" this is a version of Zariski's Main Theorem and is treated in Hartshorne, III.11.

In the statement above a simple complex-analytical, topological argument is possible. We first note that $f$ is projective, therefore proper and universally closed (also in the euclidean complex topology).

Let $Z_1,\ldots,Z_s$ be the compact connected components of the compact set $f^{-1}(q)$. Then one can choose pairwise disjoint (euclidean) open neighbourhoods $U_1,\ldots,U_s$ of $Z_1,\ldots,Z_s$. With $U = U_1 \cup \cdots \cup U_s$ the set $f(X - U) = A$ is a closed subset in $Y$, which does not contain $q$.

We now choose $V \subseteq Y - A$, open neighbourhood of $q$ and isomorphic to a ball in $\CC^r$ with $r = \dim Y$, which is possible because of $q$ being a regular point.

Let further be $B \subseteq Y$ be a closed set, such chosen, that $W = Y - B$ is a maximal open subset of $Y$ which is biregular to $f^{-1}(W) \subseteq X$ under $f$.

Because of the properties of $V$ (suitable neighbourhood of a regular point) one can assume that $V' = V - B$ is a connected open set.

With $U_i' = U_i \cap f^{-1}(V)$ we have

$$ V ' = V - B = \bigcup_{i=1}^s f(U_i' - f^{-1}(B)) $$

where $U_i' - f^{-1}(B)$ is not empty, because $\dim f^{-1}(B) < \dim X$.

But on $U_i' - f^{-1}(B)$ the map $f$ is biregular on its image, therefore 1-1 and open. So the connected set $V - B$ is a disjoint union of open sets $f(U_i' - f^{-1}(B))$. So it must be $s = 1$ and $f^{-1}(q)$ connected.

An instructive counterexample, where the assumptions are not fulfilled, is the blowup $X \to Y$ of $Y = V(y^2 - x^3 - x^2)$ over the singular point $q$, i.e. $x = y = 0$. The neighbourhood $V$ here consists of two branches, which become disconnected by removing $q$.


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