"extend a functor" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:05:59Z http://mathoverflow.net/feeds/question/52352 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52352/extend-a-functor "extend a functor" unknown 2011-01-17T22:18:08Z 2011-01-20T15:26:13Z <p>Hi,</p> <p>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.</p> <p>Thank you</p> http://mathoverflow.net/questions/52352/extend-a-functor/52359#52359 Answer by Sándor Kovács for "extend a functor" Sándor Kovács 2011-01-17T23:22:17Z 2011-01-18T03:29:10Z <p>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 <em>family</em> means so requiring "more" is an understatement).</p> <p>Here is an example to test your ideas and conditions on: Suppose $M$ is really nice, at least <em>normal</em> 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 <em>family</em> 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 $S$. </p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/52352/extend-a-functor/52634#52634 Answer by unknown for "extend a functor" unknown 2011-01-20T15:26:13Z 2011-01-20T15:26:13Z <p>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".</p>