Do Deligne-Mumford curves also have rational functions 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?
 A: 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 $p_i$.
