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Let $X$ be a topological Hausdorff space with countable basis, let $m \in \mathbf{N}$. Let us consider two definition of topological manifold with boundary.

Definition 1. $X$ is called a $m$-dimensional topological manifold with boundary iff for each $x \in X$ there exists $U \subset X$ and homeomorphism $g: U \rightarrow \overline{B_m}$ such that $x \in \operatorname{int}{U}$.

($\overline{B_m}$ denotes the closed unit ball in $\mathbf{R^m}$).

Definition 2. $X$ is called a $m$-dimensional manifold with boundary iff each $x \in X$ has an open neighbourhood which is homeomorphic either with $\mathbf{R^m}$ or with closed halfspace $\mathbf{R}_+^m = \{(x_1,...,x_m) \in \mathbf{R}^m : x_m \geq 0 \}$.

Are the above definitions equivalent (maybe with additional assumption that $X$ is compact metric space) ?


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Did you consider an annulus $\{z \in \mathbb{C} \,:\,r \leq |z| \leq R\}$ for example? This certainly is a manifold with boundary according to def. 2, however it isn't with respect to def. 1. Also, $\mathbf{R}_{+}^m$ isn't a manifold with boundary with resect to definition 1. Can you see why? – t.b. Sep 1 '11 at 12:33
@Theo If $U$ is allowed to be any (not necessarily open) neighborhood in the first definition, would the annulus satisfy it? – Dylan Moreland Sep 1 '11 at 12:38
@Dylan: You want the neighborhoods to be open (in order to ensure that the points on the boundary are mapped to the points in the boundary in the charts). Yes, if you don't require that, the annulus satisfies it but it wouldn't exhibit it as a topological manifold anymore. The issue is more that you don't want a homeomorphism onto all of the closed ball. – t.b. Sep 1 '11 at 12:45
Sorry,but I did not understand. It seems that annulus satisfies def.1. If $x$ is in interior then it is clear. If $x$ is a frontiere point, e.g. $x =(r,0)$, we take $U=\overline{B}(x, min\{r, R-r\})$. Then $x \in int U$ and $U$ is homeomorfic with closed ball. However I don't see whether $U$ is homeomorfic to closed halfspase – Richard Sep 1 '11 at 13:16
@Richard, I assume you want $U = \bar{B}(x,\min(r,R-r)) \cap X$ right? Else $U$ would contain points outside of $X$ by the above definition. How do you construct a homeomorphism then? A homeomorphism must map the boundary to the boundary, but using the induced topology by $X$ on $U$, removing a point from $\partial U$ makes it disconneted, but removing a point from $\partial B$ leaves it connected. – Willie Wong Sep 1 '11 at 13:49

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