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The motivation of this question is related to the fact that in various differencial geometry books I have seen three different criteria for chart maps. These were:

  • If $(U,\varphi)$ is a local chart, then $\varphi$ is a homeomorphism from $U$ to an open ball in $\mathbb{R}^n$.

  • If $(U,\varphi)$ is a local chart, then $\varphi$ is a homeomorphism from $U$ to an open rectangle in $\mathbb{R}^n$.

  • If $(U,\varphi)$ is a local chart, then $\varphi$ is a homeomorhpism from $U$ to an open set of $\mathbb{R}^n$.

Now, the point of charts is that their chart domains should be "trivial" topologically. From this, and some other (perhaps misunderstood by myself) remarks I have read in books, I have come under the impression that any open set of $\mathbb{R}^n$ is homeomorphic with $\mathbb{R}^n$ (this is certainly true for open balls and open rectangles, though).

This statement seems to be wrong though. For example, let $$ C=\{x^2+y^2 <r_0|\ (x,y,z)\in\mathbb{R}^3\} $$be an open infinitely long cylinder in $\mathbb{R}^3$, which is clearly an open set, so its closure $\text{cl}(C)$ is a closed set, then $$ M=\mathbb{R}^3\setminus\text{cl}(C) $$ is the complement of a closed set, so it is open.

However $M$ has nontrivial homotopy classes, in particular, loops that wind around the cylinder are not contractible. However homeomorphisms preserve homotopy, so $M$ cannot be homeomorphic to $\mathbb{R}^3$, but it is an open set of it.

Question: What is a general criterium to determine whether an open set of $\mathbb{R}^n$ is homeomorhpic to $\mathbb{R}^n$ itself?

Based on my example, I assume defining chart maps to be "just" homeomorphisms to an open subset of $\mathbb{R}^n$ is wrong then, right?

Bence Racskó
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  • It seems like this post answers your question. Some other related posts: https://math.stackexchange.com/questions/283240/how-can-i-prove-that-two-sets-are-isomorphic-short-of-finding-the-isomorphism/283260#283260, https://math.stackexchange.com/questions/1213847/are-open-sets-in-rn-homeomorphic-to-rn – Daniel Mroz Sep 16 '18 at 19:19
  • The above comment helps addressing the literal question you make. But, regarding the context, both definitions yield equivalent definitions of manifolds. There is one of them which obviously implies the other and, conversely, take notice that you can restrict the homeomorphism to a ball (since the set is assumed to be open). It ends up just being a matter of notation whether you want "charts" to be based on balls, on $\mathbb{R}^n$, or open sets in general. – Aloizio Macedo Sep 16 '18 at 19:42

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The three conditions about charts are equivalent (by perhaps different choice of "local"), because we only consider an open base for each point $p\in M$. The whole point about this "local" adjective is that you are allowed to restrict to even smaller neighbourhoods to get the condition (since in the base you can replace the old larger open set by this new smaller open set and still get a local base). There is no loss in restricting because what we ultimately considers is a maximal atlas, i.e., every chart (i.e. homeomorphism from an open subset of $M$ to an open subset of $\mathbb{R}^n$) that is compatible (i.e. transition maps are whatever class $C^0,C^1,\dots,C^\infty,C^\omega$ you are working with) with our local charts are thrown into the mix.

@DanielMroz comment and links addressed your question about how to decide if an open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ itself.

user10354138
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