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Let $X,Y$ be two quasiprojective varieties and $\phi \colon X \to Y$ a surjective morphism. Let $Z \subset Y$ a closed set such that $\phi^{-1}(Z)$ is irreducible. Prove that $Z$ is irreducible.

I do not understand what the exercise is asking. Indeed, it is well-known that a morphism is continuous (in the course I've attended we have defined morphisms as "continuous maps that preserves regular functions"); moreover, the continuous image of an irreducible set is irreducible.

So I would just answer to the exercise by saying: $\phi(\phi^{-1}(Z))=Z$ and this is true for every continuous map. Why does the exercise give me a surjective morphism? Where is my mistake?

Thank you for your help.

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The equality $\phi(\phi^{-1}(Z))=Z$ is generally only a containment $\phi(\phi^{-1}(Z))\subseteq Z$ for arbitrary $\phi$, but for $\phi$ surjective, it is equality. –  Keenan Kidwell Jul 11 '13 at 13:50
    
@KeenanKidwell Thanks for your comment. Of course you are right, so the claim is true for every continuous surjective map, am I right? I do not understand where one should use the hypothesis that $\phi$ is a morphism. –  Romeo Jul 11 '13 at 13:51
    
Dear @Romeo, Yes, it seems to me that if $f:X\rightarrow Y$ is a continuous surjective map of topological spaces and the inverse image of some set is irreducible, the set is irreducible. –  Keenan Kidwell Jul 11 '13 at 13:52
    
btw, are you sure that it is only a containment? I thought it was $\phi^{-1}\phi(Z) \subset Z$ for arbitrary $\phi$ but $\phi\phi^{-1}(Z)=Z$ for every $\phi$. –  Romeo Jul 11 '13 at 13:54
    
Dear @Romeo, I don't understand what you're saying. For any map of sets $\phi:X\rightarrow Y$, and $Z\subseteq Y$, $\phi^{-1}(\phi(Z))\subseteq Z$, and this inclusion is an equality if $\phi$ is surjective, or more generally if $Z$ is contained in the image. In general $\phi(\phi^{-1}(Z))=\phi(X)\cap Z$. –  Keenan Kidwell Jul 11 '13 at 13:55

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up vote 1 down vote accepted

Hint: Note that $\phi(\phi^{-1}(Z)) = Z$ does not hold for any map and any $Z$, you need $Z \subseteq \phi(X)$.

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But it does hold for surjective maps. –  Keenan Kidwell Jul 11 '13 at 13:49

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