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Suppose we have two smooth manifolds $M_1$ and $M_2$ and a smooth map $i:M_1 \rightarrow M_2$ that is an embedding of $M_1$ into $M_2$. Moreover we have another submanifold $N \subset M_2$ that has a non empty intersection with the embedding $i(M_1)$. Then,in what situation is the preimage set $i^{-1}(N)$ a submanifold of $M_1$?

Or in other words, what do we have to assume so that the preimage set is a submanifold?

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Do you mean "what do we have to assume so that the preimage set is a submanifold"? I assume so, but would just like to check. –  Alex Becker Dec 27 '11 at 4:45
    
Yes. I changed it. –  Mark Neuhaus Dec 27 '11 at 5:06

1 Answer 1

One hypothesis that works is that $N$ is transverse to the image of $i$, but this is not a necessary condition. For example, the preimage under the inclusion of the $x$ axis into $\mathbb R^2$ of the parabola $\{(t,t^2):t\in\mathbb R\}$ is a submanifold...

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Unfortunately the nice transversal-property is not applicable in my situation. (That is why I'm asking.) But I guess there is no other general condition and I have to do it "by hand". –  Mark Neuhaus Dec 27 '11 at 5:09
    
@Mark, you should probably explain what your situation is :) –  Mariano Suárez-Alvarez Dec 27 '11 at 5:18
    
That will not work. –  Mark Neuhaus Dec 27 '11 at 5:46
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So instead, the idea is to have people shot in the dark until they hit the hidden target. Now that will work! –  Mariano Suárez-Alvarez Dec 27 '11 at 5:51
    
The manifolds I have in mind are images of Weil functors and hence are fibered, too. –  Mark Neuhaus Dec 27 '11 at 6:23

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