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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 $Z \times_Y X$.

Edit: There are more equivalent criteria for surjectivity; see for example Tag 0B6N. It is a bit weird that the one I gave above is not listed there, especially since it is used in the proof of Tag 0BTX.

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.

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 $Z \times_Y X$.

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 $Z \times_Y X$.

Edit: There are more equivalent criteria for surjectivity; see for example Tag 0B6N. It is a bit weird that the one I gave above is not listed there, especially since it is used in the proof of Tag 0BTX.

There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of 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 $Y \to X$ 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 $Z \to Y$ 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 étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.

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 $Z \times_Y X$.