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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.

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If the fibres are connected, then yes $f_*$ is surjective, otherwise the image has finite index. The proof is as easy as 123. –  Donu Arapura Jul 20 '13 at 16:02
    
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. –  mmm Jul 20 '13 at 17:39

1 Answer 1

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 $F$.

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