Skip to main content
MathOverflow

Return to Question

added 119 characters in body; edited title
Source Link
user75933
user75933

Fibers Generalization of a morphismthe rigidity lemma in birational geometry

Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension and are connected.

If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ?

When $k=0$, that is $f(g^{-1}(z_0))$ is a point, this is true. It is known as the rigidity lemma.

Fibers of a morphism

Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension.

If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ?

Generalization of the rigidity lemma in birational geometry

Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension and are connected.

If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ?

When $k=0$, that is $f(g^{-1}(z_0))$ is a point, this is true. It is known as the rigidity lemma.

Source Link
user75933
user75933

Fibers of a morphism

Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension.

If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ?