The Riemann-Hurwitz formula implies that the projective line $\mathbb{P}^1\_K$ over any algebraically closed field $K$ is simply connected (i.e., $\pi\_1^{et}(\mathbb{P}^1\_K) = 1$; equivalently, if $\phi\colon C\to \mathbb{P}^1\_{K}$ is finite etale, then $\deg\phi=1$). 
For $K=\mathbb{C}$, this follows from the connection with topology and from the fact that the complex plane $\mathbb{A}^1\_{\mathbb{C}}$ is contractable. 
In positive characteristic the affine line is **not** simply connected due to Artin-Schreier covers.


My question is whether there is a short proof for this fact in positive characteristic? 



A short proof for simple connectedness of the projective line
You can deduce this from the classification of vector bundles on $\mathbf{P}^1$. Say $f:C \to \mathbf{P}^1$ is a connected finite etale Galois cover of degree $n$. We must show $n=1$.


The sheaf $E := f\_\* \mathcal{O}\_C$ is a rank $n$ vector bundle on $\mathbf{P}^1$, so we can write it as $E \simeq \oplus\_{i=1}^n \mathcal{O}(a\_i)$ for some integers $a\_i$. As $C$ is connected, we have $h^0(\mathbf{P}^1,E) = 1$, so we must have $a\_1 = 0$ and $a\_i < 0$ for $i > 1$ (after rearrangement). It then follows that after pullback along any finite cover $g:D \to \mathbf{P}^1$ of smooth connected curves, we still have $h^0(D, g^\* E) = 1$ as negative line bundles remain negative after pullback along finite maps, and thus cannot acquire sections.


But $f$ was a finite etale cover, so there is some $g:D \to \mathbf{P}^1$ as above with $g^\* C \to D$ is just a disjoint union of $n$ copies of $D$ mapping down (in fact, one may take $g = f$ as $f$ was Galois). But then $h^0(D, g^\*E) = h^0(D, \oplus\_{i=1}^n \mathcal{O}\_D) = n$. The only way this can happen is if $n=1$.


