Proving that a map is a morphism

Question

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.

Answer 1

It might help to identify $Sym^r(\mathbb P^1)$ with the Hilbert scheme of degree $r$ effective divisors on $\mathbb P^1$. The Hilbert scheme represents a functor, as does the space $\mathcal M_g(1,d)$, and so you can (try to) construct a map from one to the other by thinking in terms of the functors they represent.

As to why $Sym^r(C)$ is the Hilbert scheme of degree $r$ effective divisors when $C$ is a smooth projective curve, this is a (non-trivial, it has always seemed to me) exercise in making contact with reality from the somewhat more abstract world of flat families.

Answer 2

You have an answer to the 'example' version of your question already, but let me offer an answer to the "actual" question: if one is faced with two schemes $A$ and $B$, and for each $a\in A$ you have some way of constructing $f(a)\in B$, how might one check that $f$ is (or more precisely comes from) a morphism of schemes?

The answer, in many cases where $A$ is the solution to a moduli problem, is this. We're thinking of $A$ as parametrising objects $X$ (e.g. curves of genus $g$, elliptic curves with a point of order $n$ etc etc) and so for $a\in A$ you have some object $X_a$ corresponding to that point. You have a recipe that gives an element of $B$ (typically because $B$ is also the solution to a moduli problem) and you want to define the map $A\to B$ by following your nose.

But the insight is that the recipe you have, going from $A$ to $B$, might work in much more generality than you think. Let's take for example the map from the moduli space of elliptic curves plus points of order $n$, to the affine line, sending each elliptic curve to its $j$-invariant. This is "obviously" a continuous map $Y_1(n)\to{\mathbf{C}}$. But why is it a morphism of schemes?

[EDIT: I added the magic words "Weierstrass equation" to make this para correct] Well, if you go and read the definition of the $j$-invariant of an elliptic curve defined by a Weierstrass equation, then you see that if the coefficients of the Weierstrass equations are actually in a ring rather than a field, then the $j$-invariant of that curve is an element of that ring. Next one checks that $j$ is a well-defined invariant of the curve, that is, Weierstrass equations giving isomorphic curves have isomorphic $j$-invariants. But that solves your problem at a stroke! For say we have an $S$-valued point of $A$, for $S$ now any scheme. This corresponds to an elliptic curve over $S$. Now we can cover $S$ with affines such that on these affines the curve is defined by a Weierstrass equation. The $j$-invariant on these affines is a function on the affines, and uniqueness of $j$-invariants show that these functions glue (intersection of affines can be covered by affines---the usual trick) to get a well-defined function on the scheme $S$, that is, an $S$-valued point of the affine line. So for all $S$-valued points of $A$ we get an $S$-valued point of $B$ this way, just "following the definition" but applying it to the relative situation rather than the situation over fields. And the killer blow: now apply this to $S=A$, with the curve over $S$ equal to the universal curve over $A$. And there's your morphism.

[The above answer was initially too sloppy; thanks to Emerton and BCnrd for pointing this out below in the comments]

Answer 3

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 $S\to Sym^rY$.