If $X$ is a curve over a field of characteristic zero, then $X$ has a rational function, i.e., a finite morphism to the projective line.


Question. Suppose that $X$ is a Deligne-Mumford (or just algebraic) stack of dimension one over a field of characterisic zero. Does there exist a finite morphism to some $\mathbf P^1(n,m)$?


What if $X$ is assumed to be a quotient stack?



Do Deligne-Mumford curves also have rational functions
I suspect the following works. Let $\mathcal{X}$ be a proper, smooth, finite type Deligne-Mumford stack over $k$ that is one-dimensional and that has a dense open substack $U$ that is a scheme. Let $u:\mathcal{X}\to X$ be the coarse moduli space. Let $\{p\_1,\dots,p\_r\}\subset X$ be the complement of $U$. For each point $p\_i$ of $D$, let the order of the corresponding stabilizer be $n\_i$. Let $n$ be the least common multiple over every integer $n\_i$. Define $m\_i = n/n\_i$. 


Consider the divisor $D\_0$ on $X$ that is $$m\_1\underline{p}\_1 + \dots m\_r\underline{p}\_r + n\underline{q}\_1+ \dots + n\underline{q}\_s,$$
where $q\_1,\dots,q\_s$ are points different from $p\_1,\dots,p\_r$. For $s$ sufficiently large, $D\_0$ is linearly equivalent to an effective Cartier divisor $D\_\infty$ that is disjoint from $D\_0$. These two divisors together define a morphism $f:X\to \mathbb{P}^1$ such that the preimage of $\{0\}$ is $D\_0$ and the preimage of $\{\infty\}$ is $D\_\infty$. 


Now let $v:\mathbb{P}(1,n)\to \mathbb{P}^1$ be the coarse moduli space, where the unique stacky point of $\mathbb{P}(1,n)$ maps to $0$. Since $u^{-1}f^{-1}(\{0\})$ is divisible by $n$ as an effective divisor on $\mathcal{X}$, it seems to me that $f\circ u$ factors through $v$, i.e., there is a unique $1$-morphism $F:\mathcal{X}\to \mathbb{P}(1,n)$ such that $v\circ F$ is $2$-equivalent to $f\circ u$. 


Since $f$ is finite, the only thing to check is that $F$ is representable. Of course $F$ is representable away from the stacky points $p\_i$. Since $n\_i$ divides $n$, also $F$ should be representable at $p\_i$.


