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Let $M$ be an $n$-dimensional smooth compact oriented manifold with boundary which is immersed in $\mathbb{R}^n$ (codimension zero). Must $M$ be diffeomorphic to a ball with boundary (the closed unit ball in $\mathbb{R}^n$)?

Does anything change if we assume $M$ is simply-connected? or that it can be immersed in $\mathbb{R}^n$ without self-intersections?

Edit: There are (probably many) non-simply connected examples. It seems the interesting case is the simply-connected one.

Asaf Shachar
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    Consider the half open interval? – Pedro Jun 09 '16 at 17:20
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    What about solid torus – Anubhav Mukherjee Jun 09 '16 at 17:20
  • @PedroTamaroff Thanks, nice idea... Your example (the half open interval) is simply connected and of course can be immersed in $\mathbb{R}$ without self-intersections. Can you think on a simply-connected example in higher dimensions? (Also, I was actually interested in compact examples but forgot to mention this) – Asaf Shachar Jun 09 '16 at 17:39
  • compact one does not exists: http://math.stackexchange.com/questions/1519028/torus-cannot-be-embedded-in-mathbb-r2 –  Jun 09 '16 at 21:27
  • @JohnMa I am not sure I understand your comment. Isn't the closed unit ball an example for a manifold with boundary which is immersed in $\mathbb{R}^n$ (via the standard inclusion). I think the argument you are referring to (about the open+closed+connectedness issue) "fails" at the boundary. Am I wrong? – Asaf Shachar Jun 09 '16 at 22:50
  • Clearly I ignored the word "with boundary" in the question. –  Jun 10 '16 at 00:13
  • Anyway you can just pick any open set in $\mathbb R^n$ with smooth boundary. Even when you assume that it is simply connected, there are lots of counterexamples. –  Jun 10 '16 at 00:14
  • As for more interesting examples, each open stably parallelizable $n$-manifold admits an immersion in $R^n$. For instance, every open orientable 3-dimensional manifold immerses in $R^3$. – Moishe Kohan Jun 10 '16 at 14:19

2 Answers2

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Consider a closed annulus in $\mathbb{R}^2 $. It is not even homeomorphic to a closed disk.

Hmm.
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Consider any emebdded compact submanifold of dimanions $k$ in $\bf R^n$, say $V$. Then a tubular neighbourhood $W$ of $V$ is simply connected, but not a ball (it has the same homotopy type than $V$. For instance if $V=S^2$ is the unit two-sphere in $\bf R^3$ the set $1/2\leq x^2+y^2+z^2 \leq 1+1/2$ is a compact simply connected manifold with the same homotopy type than a sphere, hence simply connected and not contractible. If the codimension of $V$ is at least two, the boundary of $W$ is connected.

Thomas
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