It is well-known that the closed $n$-ball has Euler characteristic $1$. Is it true that every closed (i.e., compact), connected $n$-dimensional submanifold (with boundary) of $\mathbb R^n$ having Euler characteristic $1$ must be diffeomorphic to the closed $n$-ball?
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1$\begingroup$ I guess "closed" means "without boundary" in usual English conventions. $\endgroup$– YCorCommented Apr 19, 2021 at 12:56
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$\begingroup$ This question is more appropriate for Math Stack Exchange. However, if you want to ask there, make sure, include your own attempts to solve the problem. As a hint, think if you know any surfaces (besides the disk) which have Euler characteristic 1. $\endgroup$– Moishe KohanCommented Apr 19, 2021 at 16:15
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1 Answer
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No.
Take any finite simplicial complex $K$, find an $n$ for which $K$ embeds (piecewise linearly, even!) in $\mathbb{R}^n$. Then for sufficiently small $\varepsilon > 0$, the $\varepsilon$-neighborhood of $K$ will be a compact manifold with boundary, and homotopy equivalent to $K$. So there are plenty of examples.
answered Apr 19, 2021 at 12:33