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Let $X \to Y$ be a projective morphism. So, this map factors through a closed immersion $i$ of $X$ into $\mathbb{P}^n \times Y$ for some $n$ followed by the projection map to $Y$. When is it possible to define a map $\phi$ from $Y$ to $\mathbb{P}^n \times Y$ such that the image of $\phi$ is contained in the image of the closed immersion $i$ such that $\phi$ composed with the projection map is identity? When can we say that $\phi$ is itself a closed immersion.

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Well, a trivial condition is that $dim(X)\geq dim(Y)$ and if equal then $X=Y$. I don't see any other very general conditions. –  IMeasy Feb 7 '13 at 17:13
    
@IMeasy: This seems to a very general condition. Could you please elaborate or suggest some reference for your answer. –  Naga Venkata Feb 7 '13 at 17:29
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IMeasy: I guess you're saying that these are trivial necessary conditions, but I think the situation is a bit more subtle than you claim in the case of equal dimension. First of all, one could do something stupid like have X be two disjoint copies of Y. So maybe you want to restrict to X connected. But in that case, one could take Y to be Spec(k), and X to be Spec(k[t]/t^2). –  Artie Prendergast-Smith Feb 7 '13 at 18:58
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I do not think there are general conditions for existence of $\phi$. Do you have some particular case in mind? If $\phi$ exists, it is clearly a closed immersion, since it is the graph of a morphism $Y\to\mathbb{P}^n$. –  Mohan Feb 7 '13 at 19:33
    
Also, I agree with Mohan that the question is too broad to expect a reasonable answer. It would be better to ask about a particular case. –  Artie Prendergast-Smith Feb 7 '13 at 19:53

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