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?
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). $\endgroup$ – Jason Starr Nov 18 '15 at 10:38

$\begingroup$ You're absolutely right; that was very bad. I've added finiteness conditions throughout. $\endgroup$ – R. van Dobben de Bruyn Nov 18 '15 at 13:32

$\begingroup$ It was just a typo, it was not very bad :) $\endgroup$ – Jason Starr Nov 18 '15 at 13:43