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Assume that $f: X \rightarrow Y$ is a smooth map between two smooth manifolds.

Must there exist a smooth manifold $Z$, a submersion $g:X \rightarrow Z$, and an immersion $h:Z \rightarrow Y$ such that $f = h \circ g?$ Why or why not?

okipik
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    No. Take a smooth 1-1 map that is a topological embedding $f : \mathbb R \to \mathbb R^2$ whose image is not a smooth manifold. – Ryan Budney Aug 13 '13 at 17:26
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    More generally, the derivative does not need to have constant rank. For your question to have a positive answer, the derivative would have to have constant rank. – Ryan Budney Aug 13 '13 at 17:30
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    $f(x)=x^2$ for example – Ryan Budney Aug 13 '13 at 17:37
  • @RyanBudney Pardon the possibly stupid question, but why does the converse hold (constant rank implies composition)? It's obvious in coordinates, but how does one get a global composition? Thanks! – Alex Youcis Aug 13 '13 at 18:01
  • I did not say an if and only if statement, only a one-way implication. – Ryan Budney Aug 13 '13 at 18:03
  • @RyanBudney Do you know a characterization of the maps which are the composition of an immersion and a submersion? What about the composition of an embedding and a submersion? – Alex Youcis Aug 13 '13 at 18:14
  • In generality there's not much that wouldn't be just some relabelling of the words "immersion" and "submersion". But if the fibres of the map are compact and if the derivative has constant rank, then that would be enough. I think this is a theorem that has a name (which I always forget), but it's just a basic local-to-global argument. – Ryan Budney Aug 13 '13 at 21:53

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