# “extend a functor”

Hi,

I have probably a basic question. I have a functor $F: Sch \rightarrow Set$, an algebraic stack $M$ with a "universal family" $G\rightarrow M$ and a representability property like this: for every family of group schemes $L\rightarrow S$ where $S$ is a normal scheme there exists a unique map $S\rightarrow M$ such that $L\rightarrow S$ is the pullback of $G\rightarrow M$ in a unique way. The question are: can the assumption on normality of $S$ be relaxed? More precicely, does exists any "obstruction" theory which says when this is possible at least in good situations for $M$? If we need we can require that $M$ has as good properties as we want.

Thank you

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Why did you mention the functor F? It doesn't seem relevant to the question at all. I don't think the question warrants the category-theory tag. – David Roberts Jan 18 '11 at 0:30
I doubt that you really mean "every family of group schemes", without further restriction. Next, what do you mean by "is the pullback in a unique way"? I understand this as "there is a unique isomorphism of $L$ with the pullback", which implies that $L$ has no nontrivial automorphisms! – Laurent Moret-Bailly Jan 18 '11 at 7:58
ok. The statement was not correct in details because what I want to know is that if there are good "settings" for $G\rightarrow M$ and $L\rightarrow S$ in order to solve the "descent" from normal scheme to schemes which are less "good". – unknown Jan 20 '11 at 15:26

## 1 Answer

It seems to me that without requiring more from the family $L\to S$ this will not hold. (Well, you didn't really say what family means so requiring "more" is an understatement).

Here is an example to test your ideas and conditions on: Suppose $M$ is really nice, at least normal and suppose there exists a morphism $\alpha:M\to S$ which is one-to-one on closed points but not an isomorphism. Say $M=\mathbb A^1$, $S$ is a cuspidal cubic and $\alpha$ is the normalization. Now consider the family on $S$ obtained by composing $\alpha$ with the morphism $G\to M$. So you have essentially the same family, at least the same fibers (kinda) just a little cusp-ed at some points of $S$.

Now if this is an admissible family in your situation (I guess it may not be as the fiber over the cusp will be a multiple fiber and you probably disallow that, but who knows), then you have a problem: If this family is to be pulled-back from $M$, then the desired morphism should be an inverse to $\alpha$ (at least point-wise), but we know that there is no such map as a non-normal point cannot dominate a normal one with degree one.

On the other hand, what you can certainly do in your original situation is to pull-back the family $L$ to the normalization of $S$ and obtain your map to $M$ that exhibits it as a pull-back. Then you can try to analyze the situation and see if this morphism from the normalization of $S$ would factor through $S$. The main question is whether a crooked family as above is possible.

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