Let $f:X \to Y$ be a morphism of schemes. I am interested in sufficient conditions on $f$ which would ensure that the induced map $\pi_1^{et}(X) \to \pi_1^{et}(Y)$ of etale fundamental groups is surjective. According to this question https://math.stackexchange.com/questions/491305/etalefundamentalgroup SGA 1 X Corollary 1.4 says that it is true when $f$ is proper, separable, $Y$ is connected, and $f_*(\mathcal O_X)=\mathcal O_Y$ (in particular, $f$ has geometrically connected fibers). Specifically I am interested if the condition that $f$ be a (not necessarily proper) Zariski locally trivial fibration with connected fibers (where $X$ and $Y$ are smooth algebraic varieties) is sufficient for the map $\pi_1^{et}(X) \to \pi_1^{et}(Y)$ induced by $f$ to be surjective?
1 Answer
There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of finite étale covers of $X$ is equivalent to the category $\pi_1(X)\text{ }\operatorname{Set}_f$ of finite sets with a continuous $\pi_1(X)$action. Under this correspondence, the $Y \to X$ finite étale with $Y$ connected correspond to the connected $\pi_1(X)$sets $S$ (i.e. $\pi_1(X)$ acts transitively on $S$).
Lemma. Assume $X$, $Y$ connected, and $f \colon X \to Y$ a morphism. Then the induced morphism $\pi_1(f) \colon \pi_1(X) \to \pi_1(Y)$ is surjective if and only if for every $Z \to Y$ finite étale with $Z$ connected, the pullback $Z_X \to X$ is connected.
Proof. If $\pi_1(f)$ is surjective, then clearly any connected $\pi_1(Y)$set is connected as $\pi_1(X)$set. Conversely, if $\pi_1(f)$ is not surjective, then some $\gamma \in \pi_1(Y)$ is not in the image. Since fundamental groups are profinite, the image of $\pi_1(f)$ is closed, so the image of $\pi_1(f)$ misses some open neighbourhood of $\gamma$. Thus, there exists an open subgroup $U \subseteq \pi_1(Y)$ such that $$\gamma U \cap \operatorname{im} \pi_1(f) = \varnothing.$$ Then the finite $\pi_1(Y)$set $S = \pi_1(Y)/U$ is not connected as $\pi_1(X)$set. But it is clearly connected as $\pi_1(Y)$set. $\square$
To apply this to the specific geometric setting you are interested in, just note that if $f \colon X \to Y$ has connected geometric fibres, then the same holds for the base change to any finite étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.
Remark. There are more equivalent criteria for surjectivity; see for example Tag 0B6N. The one I gave above is amongst the ones listed, but this was not the case at the time of writing; hence my writing out the proof. My proof above was originally part of the proof of Tag 0BTX.

$\begingroup$ In the statement of your lemma, I think you should clarify that also $Z\to Y$ is finite (i.e., proper). $\endgroup$ Nov 18, 2015 at 10:38

$\begingroup$ You're absolutely right; that was very bad. I've added finiteness conditions throughout. $\endgroup$ Nov 18, 2015 at 13:32

$\begingroup$ It was just a typo, it was not very bad :) $\endgroup$ Nov 18, 2015 at 13:43

$\begingroup$ If $X$ and $Y$ are quasiprojective varieties over $\mathbb{C}$, does this lemma holds for the topological fundamental groups? Namely, under the same assumptions for $f$ and $X$, $Y$, is $\pi^{\rm top}_1(X(\mathbb{C}))\to\pi_1^{\rm top}(Y(\mathbb{C}))$ surjective? $\endgroup$ Jul 29, 2023 at 10:21

$\begingroup$ @HiggsBoson hmm, if you only pose a condition on finite coverings, you can not really say something about the topological fundamental group; only its profinite completion. I think topological fundamental groups of varieties need not be residually finite in general, so there is no formal argument. On the other hand, if you have the connectedness condition on all (possibly infinite) topological coverings, then you can deduce surjectivity in much the same way (using 'Galois theory for covering spaces'). $\endgroup$ Jul 29, 2023 at 21:37