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Let $G_1, ..., G_s : \mathbb{R}^s \to \mathbb{R}^s$ be diffeomorphisms on an open unit ball $B \subset \mathbb{R}^s$, centered at $0$, onto their images. Let $m_1, .., m_s$ be integers not all $0$, and let $F = m_1 G_1 + ... + m_s G_s$. Suppose we know that the determinant of the Jacobian matrix of $F$ at $x$ is non-zero for all $x \in B$.

  1. Does it then follow that $F$ is still a diffeomorphism on $B$?

  2. If $F$ is not a diffeomorphism on $B$, is it possible to show that $$ \# F^{-1}(y) < C $$ for all $y$ in $F(B)$ for some positive $C$?

Any comments are appreciated!

Ps edits had been made based on comments

Johnny T.
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  • I think 1 admits an easy counterexample. Take $F = G - G = 0$ for any diffeo $G$. – Charles Hudgins Mar 26 '21 at 22:33
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    I was hoping that the assumption that the Jacobian of $F$ at $0$ is non-zero would remove easy counterexamples... – Johnny T. Mar 26 '21 at 22:35
  • Is that in the OP? If you know the jacobian is non-zero, you have that $F$ is a local diffeo and hence has discrete fibers. The question is whether there's a uniform, finite bound on the fibers, right? – Charles Hudgins Mar 26 '21 at 22:38
  • Sorry I meant to write the determinant of the Jacobian matrix of $F$ is non-zero. Typos had been corrected. I hope it's clearer now, thank you! – Johnny T. Mar 26 '21 at 23:01
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    I liked the previous version of the question more. You are right, the Jacobian condition at a single point rules out the trivial counterexamples. For example, for $F=G-G$ we have $dF(0)=0$, obviously. You posed a good question. This version is also interesting, though. – Giuseppe Negro Mar 26 '21 at 23:01
  • thank you! I am still curious about the previous version, but I thought I should ask a more easier question first and try thinking about it – Johnny T. Mar 26 '21 at 23:03
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    The $m_i$ are irrelevant because $m_iG_i$ is a diffeomorphism if $G_i$ is one. – Paul Frost Mar 28 '21 at 00:00
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    @Paul, also if (1) holds for two diffeos, it holds for $n$, by induction. – Ruy Mar 28 '21 at 00:43
  • It looks like (1) holds if $F$ is proper and the image is simply connected by a result by Hadamard https://math.stackexchange.com/questions/41551/global-invertibility-of-a-map-mathbbrn-to-mathbbrn-from-everywhere-loc?noredirect=1&lq=1 Although it's not clear to me if these conditions will be satisfied automatically in my situation... – Johnny T. Mar 28 '21 at 15:26

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