4
$\begingroup$

Let $f:X\to B$ be a smooth projective morphism of complex algebraic varieties.

If $f$ is of relative dimension zero, i.e., $f$ is a finite etale cover, then the image of the topological fundamental group of $X$ in the topological fundamental group $\pi_1(B)$ of $B$ is of finite index.

What is $f$ is of relative dimension one? What are the properties of the morphism $f_*:\pi_1(X)\to \pi_1(B)$? Is it always surjective?

What if $f$ is of relative dimension $n$?

I am especially interested in the case where $f$ is non-isotrivial and $B$ is a curve.

$\endgroup$
  • 3
    $\begingroup$ If the fibres are connected, then yes $f_*$ is surjective, otherwise the image has finite index. The proof is as easy as 123. $\endgroup$ – Donu Arapura Jul 20 '13 at 16:02
  • $\begingroup$ relevant facts(references) are collected at pages 15-17 of arxiv.org/abs/0905.1377 ; there should be an answer to your question as well. the goal there was to prove that the universal analytic cover of an projective variety has a Zariski-like topology where projections are closed. $\endgroup$ – mmm Jul 20 '13 at 17:39
2
$\begingroup$

A smooth projective map of complex varieties is a proper submersion, and thus a fibration by Ehresmann's theorem. In particular, the long exact sequence of homotopy groups gives an exact sequence (of pointed sets) $$\pi_1(X)\to \pi_1(B)\to \pi_0(F)\to 1$$ where $F$ is a fiber, where we suppose $X$ and $B$ are connected to avoid issues with a choice of basepoint. So as Donu Arapura remarks, the index of the image may be computed in terms of the number of connected components the fiber $F$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.