Let $k$ be an algebraically closed field, it is well known that $\mathbb P^1$ is simply connected, but how about smooth projective surfaces $X$ with a smooth morphism to $\Bbb P^1$?
Except the case $X$ is a product of curves or a projective bundle like Hirzebruch surfaces, could we classify all such $X \rightarrow \Bbb P^1$ ? What about higher dimensional cases?
Motivation: Shafarevich conjecture over function field.
Let $k$ be an algebraically closed field.
Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).
Theorem. The morphism $f$ is isotrivial.
Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED
[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.
Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.)
Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.
If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.
If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below.
If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.
