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I know that the maps between manifolds without boundary which are both a submersion and an immersion are exactly the local diffeomorphisms, and I wonder if it can happen that a map from a manifold without boundary to a manifold with boundary that is both a submersion and an immersion be not a local diffeomorphism. If it is the case, do you know any example?

Thank you in advance for any help.

Lazarus Frost
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    This cannot happen. Try showing that a submersion between manifolds with boundary maps interior points to interior points. Thus, in case the domain doesn't have boundary, you can replace the codomain by its interior and apply your first result. – Thorgott Mar 15 '23 at 16:43
  • @Thorgott I've reduced your approach to proving that the image of a smooth submersion $U \subset \mathbb{R}^n \rightarrow V \subset \mathbb{H}^m$ can't intersect the boundary of $\mathbb{H}^m$, but I'm stuck here. Could you give some guidance? – Lazarus Frost Mar 16 '23 at 01:23
  • Being a submersion means that the differential is surjective. That is, if $x \mapsto y$, then every vector at $y$ is the image of some vector at $x$. If $y$ is a boundary point, then there are some directions from $y$ that you cannot go. But if $x$ is an interior point, then you can go in every direction from $x$. Do you see the problem? – Paul Sinclair Mar 16 '23 at 19:59
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    Assume $g\colon\mathbb{R}^n\rightarrow\mathbb{H}^n$ is such that $g(0)=0$. It suffices to show that $dg\vert_0$ cannot be surjective. If $t_n\colon\mathbb{H}^n\rightarrow[0,\infty)$ is projection on the last coordinate, then $t_n\circ g$ has a (global) minimum at $0$. What does that say about its derivatives and the Jacobian of $g$ at $0$? – Thorgott Mar 17 '23 at 02:37
  • @PaulSinclair I have the same intuition. However, one can prove that the differential of the inclusion $\mathbb{H}^n \hookrightarrow \mathbb{R}^n$ is an isomorphism at every boundary point; namely, at a boundary point of $\mathbb{H}^n$, one has no fewer directions than that one have in $\mathbb{R}^n$; or is this a wrong interpretation of that result? – Lazarus Frost Mar 17 '23 at 15:19
  • @Thorgott Does it mean that the last row of such a Jacobian is a row of zeros, so it can't be of full rank? – Lazarus Frost Mar 17 '23 at 15:34
  • You got it.${}{}$ – Thorgott Mar 17 '23 at 17:01
  • That one can prove the differential of a map from a manifold with boundary $\to$ a manifold without boundary is an isomorphism on boundary points is immaterial when the question is about a map from a manifold without boundary $\to$ a manifold with boundary. When $g$ carries a point in which curves can pass through in any direction, and thus can have any vector in the tangent space as their tangent, to a point where curves cannot pass through in some directions, because that would take them out of the manifold, then those directions cannot be in the image of $dg$. – Paul Sinclair Mar 17 '23 at 17:51

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