TITLE: Proving that a map is a morphism
QUESTION [5 upvotes]: Example : Consider the (open, not compactified) moduli space of stable maps $ \mathcal M_g(1,d)$ of maps of smooth curves of genus $ g$ to $\mathbb P^1$. To each map
we associate its branch divisor, which is an element of $ Sym^r(\mathbb P^1)$. Then we should have a morphism from $ \mathcal M_g(1,d)$ to $ Sym^r(\mathbb P^1)$. How do
we prove this?
In general, if we have a family of objects $A$ and for each $ a \in A$ we can
choose an element $ b \in B$ that "depends continuously" on $ a$. How do we prove
that we have established a morphism $A \to B$? What is the general method to do this?
Note about the example: Extending the morphism to the Kontsevich compactification of the moduli space
was the main objective of a paper by Fantechi-Pandharipande. But I couldn't filter out
the proof that the map is a morphism, or the proof wasn't there.

REPLY [6 votes]: Emerton has already answered, but let me summarize all the steps:
1) to a perfect torsion complex in the derived category associate a Cartier divisor;
2) to every sequence of morphisms $C\to X\to S$ satisfying suitable assumptions associate a perfect torsion complex, hence a Cartier divisor;
3) when $C\to X\to S$ is a family of stable maps to a smooth curve $Y$ (thus $X=Y\times S$), the divisor obtained on $X$ is effective and commutes with base change.
4) since $M_g(X,d)$ is the stack of stable maps, i.e. it represents a pseudofunctor (or, if you prefer, its coarse moduli space corepresents a functor), to define a morphism from it to a scheme $T$ means to define for every family over $S$ a mor $S\to T$ commuting with base change.
5) as Emerton wrote, $Hilb^rY=Sym^rY$. Hence an effective Cartier divisor on $X=S\times Y$ which is of degree $r$ on every fiber of $X\to S$ defines a morphism $S\to Sym^rY$.