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Sep 11, 2013 at 15:53 vote accept Keesjan
Sep 11, 2013 at 15:53 comment added Keesjan Thnx for this. My silly question was aiming at the statement that "if $\mathcal X$ has stacky points, then no morphism $\mathcal X\to \mathbf P^1$ is representable". Thank you very much for your answer. :)
Sep 11, 2013 at 15:37 comment added Jason Starr @Keesjan: You know what you were asking about better than I do. However, my interpretation of a "finite morphism" of stacks is a morphism that is representable by finite morphisms. If $\mathcal{X}$ has stacky points, then no morphism $\mathcal{X}\to \mathbb{P}^1$ is representable. The construction above produces a representable morphism $\mathcal{X}\to \mathbb{P}(1,n)$.
Sep 11, 2013 at 13:00 comment added Keesjan My apologies for the unusual etiquette. My follow-up question is actually related to your answer. In fact, you construct a finite morphism $\mathcal X\to X \mathbf P^1$ which factors through some $\mathbf P^1(n,m)$. So this is probably very stupid, but why bother finding a factorization? Doesn't the morphism $\mathcal X\to X \mathbf P^1$ already answer the question? Is it not a "rational function"? Hope I'm making sense...
Sep 11, 2013 at 12:20 comment added Jason Starr @Keesjan: Usually it is best to ask a follow-up question as a separate question on MO. I am not sure I understand your follow-up question: if I take $n=m=1$, then the morphism $v:\mathbb{P}(n,m)\to \mathbb{P}^1$ is an isomorphism.
Sep 11, 2013 at 8:26 comment added Keesjan Hi, is it 'easy' to write down an example of a morphism $\mathcal X \to X \to \mathbf P^1$ which does not factor through a 1-morphism $F:\mathcal X \to \mathbf P^1(n,m)$ for any $n,m$?
S Sep 10, 2013 at 15:34 history answered Jason Starr CC BY-SA 3.0
S Sep 10, 2013 at 15:34 history made wiki Post Made Community Wiki by Jason Starr