Existence of covers for given genus and degree. Suppose we are given a genus $g$ and degree $n$; under which circumstances is there a curve $C$ of that genus admitting a map of degree $n$ to $\mathbb{P}^1$?
For example: If $n = 2g-1$ there is no such curve. More precisely any line bundle $L$ on a genus $g$ curve of degree $2g - 1$ has a fixed point by Riemann-Roch: $$h^0(L) - h^0(K_C - L) = h^0 (L) = 2g - 1 - g + 1 = g$$ and for $p = L - K_C$ we have: $$h^0(L- p) - h^0(K_C - L + p) = h^0 (L - p) - 1 = 2g - 2 - g + 1 = g - 1$$ Thus $h^0(L- p) = h^0(L)$ and $p$ a fixed point. 
On the other hand, by the same calculation, if $n \geq 2g$ every line bundle of this degree will do. For $n = 2g - 2$ the canonical bundle will do. What about $n < 2g - 2$, are there any more pairs $(g, n)$, for which such a cover doesn't exist? 
Edit: Ok, so the above argument does not work, since $p = L - K_C$ requires $L - K_C$ to be effective, which in general is not the case. So, are there any such pairs $(g, n)$ for $n > 1$? 
 A: Your question is essentially about Brill - Noether theory. The existence theorem of BN - theory tells us that if $C$ is a curve of genus $g$ and  $g-(r+1)(g-d+r)\geq 0$ then there exists a $g^r_d$ (i.e a degree $d$ line bundle with a choice of a subspace $V\subseteq H^0(L)$ of dimension $r+1$) on $C$. So if you put $r=1$ above, you will see that a genus $g$ curve admits a map to $\mathbb{P}^1$ of degree $d$ as long as $d\geq \frac{g+2}{2}$. Moreover BN - theory tells us that this is the generic situation, meaning that a general curve has no $d:1$ map to $\mathbb{P}^1$ if $d< \frac{g+2}{2}$.
Besides, the argumentation in your example is not correct because you are assuming that for any line bundle of degree $2g-1$ you have $h^0(L-K_C)\geq 1$, which is not always true. In fact, such line bundles are all of the form $K_C+p$ for some $p\in C$, hence they constitute a $1$ dimensional family inside $Pic^{2g-1}(C)$, which is $g$ dimensional.
A: For $g = 2$ and $n = 2g - 1 = 3$ there are curves like this. Namely $C$ defined as the projective nonsingular model of $y^3 = x(x - 1)(x - 2)^2(x - 3)^2$ has genus $2$ and comes with a degree $3$ morphism to $\mathbf{P}^1$.
I do agree that with $g = 1$ and $n = 2g - 1 = 1$ there is no such curve.
The question you are asking is decidable. I suggest reading about Riemann surfaces and the Riemann-Hilbert correspondence.
A: Ok, so I guess there is always such a cover for $(g,n)$, $n > 1$: Choose $2g - 2 + 2n$ points on $\mathbb{P}^1$ and call the set of those points $S$. Then a n-sheeted top. cover of $\mathbb{P}^1$ is given by an element $\sigma_i$ in the permutation group on $n$ elements for every point of $S$, s.t. $\sigma_1 \cdot ... \cdot \sigma_{2g - 2 + 2n} = id$ and the subgroup generated by them acts transitively on the set of $n$ elements. This we can do by choosing transpositions over every point. By the Riemann existence theorem this extends to a covering of Riemann surfaces and by the Hurwitz formula the resulting Riemann surface has the required genus.    
