In his chapter about Hurwitz' theorem for curves, Hartshorne shows that $\mathbb{P}^1$ is simply connected, i.e. every finite étale morphism $X \to \mathbb{P}^1$ is a finite disjoint union of $\mathbb{P}^1$s. In an exercise the reader is invited to show that $\mathbb{P}^n$ is simply connected, using the result for $\mathbb{P}^1$.


I have no idea how to do this. Perhaps someone can give a hint? There are closed immersions $\mathbb{P}^1 \to \mathbb{P}^n$, along which we may pull back a finite étale morphism, but the trivializations don't have to coindice ... perhaps we can resolve this using cohomology theory? I'm a bit confused since $\mathbb{P}^n$ is $n$-dimensional, but this is in Hartshorne's chapter about curves. I don't want to use the more advanced material of SGA.



$\mathbb{P}^n$ is simply connected
Here is a sketch of an argument which directly uses simple connectedness of $\mathbb P^1$, and is related to the simple connectedness of rationally connected smooth varieties mentioned by Sandor in one of his answers. 


The idea is to treat the $\mathbb P^1$s in $\mathbb P^n$ as analogous to arcs in a topological space, and to make a lifting argument (just as one does in the basic topological theory of covering spaces).


Let $Y \to \mathbb P^n$ be a finite etale map. Fix a base points $x \in \mathbb P^n$ and
a point $y \in Y$ lying over $X$. If $x' \in \mathbb P^n \setminus \{x\}$, there is a unique
line $L$ joining $x$ and $x'$. The preimage of $L$ is a disjoint union of curves $L'$, each mapping isomorphically to $L$ (by simple connectedness of $\mathbb P^1$), and we can choose a unique $L'$ containing $y$. Now let $y'$ be the point of $L'$ lying over
$x'$.


The map $x' \mapsto y'$ (and of course mapping our original point $x$ to $y$) gives a section to the given map $Y\to \mathbb P^n$, which is what we wanted.


Added: Here is one explanation of why the map $x' \mapsto y'$ is algebraic. Let $\pi:Y \to \mathbb P^n$ be our given etale map. First note that
$x' \mapsto \pi^{-1}(L)$ (where $L$ is the line joining $x$ and $x'$, as above) is a morphism from $\mathbb P^n \setminus \{x\}$ to the Hilbert scheme of $Y$. Now picking out the connected component $L'$ of $\pi^{-1}(x')$ containing $y$ is a morphism from our given locally closed subset of the Hilbert scheme to the Hilbert scheme, and so altogether we
see that $x' \mapsto L'$ is a morphism. Finally, mapping $L'$ to $x'$ (which can be described as forming the intersection $L' \cap \pi^{-1}(x')$) is again a morphism. So altogether we have a section $\mathbb P^n \setminus\{x\} \to Y$. One way to show that
this extends as a section over all of $\mathbb P^n$ (by sending $x$ to $y$) is just
to repeat the whole process for a different choice of $x$, and glue the two resulting
sections. 


