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As described here (and as I always thought was the most general definition of boundary), a possible definition of the boundary of a subset $S$ of a topological space $X$ is $\partial S = \overline S \backslash \mathrm{int}(S)$.

I am reading Allen Hatcher's book on algebraic topology, and they often refer to "the" boundary of a topological space, for instance by saying that $\partial D^2 = S^1$. How is this notion of boundary uniquely defined? Because if I take an injection of $D^2$ into $\mathbb R^2$, I'm going to get $S^1$, but if I take the injection of $D^2$ into $\mathbb R^3$, then $D^2$ becomes its own boundary. I know that the $\partial$ is an important operation in topology so I'm trying to figure it out, and I feel like it doesn't quite make sense here.

So can anyone describe precisely what is meant by $\partial X$ when $X$ is a topological space? Or at least explain what is meant by that in particular contexts where it is used.

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  • $\begingroup$ This is not the topological notion of boundary as you observed. The context you're probably looking for is that of "topological manifolds with boundary". $\endgroup$ Sep 1 '13 at 7:05
  • $\begingroup$ @Anthony : I figured it wasn't the definition of the boundary of a topological space... but I guess I'll look up that topological manifold boundary thing! $\endgroup$ Sep 1 '13 at 7:09
  • $\begingroup$ As you have observed, $\overline{S}\setminus int(S)$ is not intrinsic to $S$, but also depends on $X$. If you view $D^2$ as a subset of itself, then its boundary becomes empty. Therefore in the context of manifolds a different definition is needed. +1 for a good question. $\endgroup$ Sep 1 '13 at 7:19
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    $\begingroup$ For geodesic metric spaces, there is also a notion of boundary obtained by considering equivalence classes of geodesic rays 'tending to infinity'. $\endgroup$
    – user68316
    Sep 1 '13 at 7:23
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    $\begingroup$ A more general definition would be to talk about the local homeomorphism type at a point in a topological space. For a manifold with boundary there are only two such points, interior points and boundary points. This has the advantage of making sense in arbitrary topological spaces. $\endgroup$ Sep 3 '13 at 18:30
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The notion of boundary that you are looking for comes from the definition of topological manifolds with boundary. As opposed to a regular manifold $X$, a manifold with boundary has the property that each point in $X$ has an open neighborhood which is homeomorphic to an open set in the euclidean half space $\mathbb{R}_+^n=\{(x_1,\dots,x_n)\in\mathbb{R^n}:x_n\ge0\}$. Thus we then define $\partial X$ to be the points which when mapped to $\mathbb{R}_+^n$ have $x_n=0$.

This definition has the benefit that an embedding of $X$ into some other space does not change $\partial X$. Thus $\partial D^2=S^1$ irregardless of whether you view it as living in $\mathbb{R}^2$ or in $\mathbb{R}^3$.

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    $\begingroup$ When you say "all points in $X$", you mean "each point of $X$ has some neighborhood such that"? $\endgroup$ Sep 1 '13 at 9:57
  • $\begingroup$ @PatrickDaSilva thanks! I have fixed the error now $\endgroup$
    – E.O.
    Sep 1 '13 at 12:34
  • $\begingroup$ This seems to restrict to a very particular kind of space, i.e. those whose boundary is $(n-1)$-dimensional (i.e. in the sense that its boundary has the property that each point has an open neighborhood that maps homeomorphically to an open set in $\mathbb R^{n-1}_+$. I would think, for instance, that in this "manifold with boundary" context, if you see the letter $P$ as a topological space (where the hole in the $P$ is considered as its interior, i.e. $\pi_1(P) = 0$), then the tail of the $P$ would be a part of its boundary, which does not seem the case with your definition. $\endgroup$ Sep 1 '13 at 13:47
  • $\begingroup$ I'm asking because I see this chapter with cell-complexes in my book where they glue a bunch of stuff together and its possible to glue a line to a disk (to get the letter $P$ for instance). How does one go around that in the literature? Or is it even a problem? $\endgroup$ Sep 1 '13 at 13:48
  • $\begingroup$ @PatrickDaSilva you correct in that the boundary is necessarily $(n-1)$-dimensional. For your example though, $P$ is not a manifold. When you map the tail to $\mathbb{R}^2$, it will be a line. But lines are not open in $\mathbb{R}^2$ so $P$ is not a manifold. $\endgroup$
    – E.O.
    Sep 1 '13 at 22:20
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The concept of boundary can be extended to the (regular) CW-complexes, as suggested here.

The boundary of a (regular) CW-complex $X$ is : $$\partial X := \overline{\bigcup_{n \ge 0}(\bigcup_{c \in \text{n-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{n-cells}} (\partial c \cap \partial c'))}$$

Definition : The notation "$n$-cells" above, is the set of closed $n$-cells.


Example : Let $X$ be a topological space with the following simplicial complex structure :
enter image description here

All the sets :

  • $0$-cells $=\{ A,B,C,D,E,F \}$
  • $1$-cells $=\{ [A,B],[B,C],[C,D], [D,E],[E,A],[A,F]... \}$
  • $2$-cells $=\{ [A,B,F],[B,C,F],[C,D,F], [D,E,F],[E,A,F]\}$

Now :

  • $\partial A = \partial B = ... = \partial F = \emptyset$
  • $\partial [A,B] = \{A,B \}$ , $\partial [B,C] = \{B,C \}$ , ....
  • $\partial [A,B,F] = [A,B] \cup [B,F] \cup [A,F] $, $\partial [B,C,F] = [B,C] \cup [C,F] \cup [B,F] $, ...

So :

  • $(\bigcup_{c \in \text{0-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{0-cells}} (\partial c \cap \partial c')) = \emptyset$
  • $(\bigcup_{c \in \text{1-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{1-cells}} (\partial c \cap \partial c')) = \emptyset$
  • $(\bigcup_{c \in \text{2-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{2-cells}} (\partial c \cap \partial c')) = (A,B) \cup (B,C) \cup (C,D) \cup (D,E) \cup (E,A)$

Conclusion : $\partial X = [A,B] \cup [B,C] \cup [C,D] \cup [D,E] \cup [E,A]$


Questions : Let $X$ be a topological space admitting a (regular) CW-complex structure :

  • Does $\partial X$ depend on the choice (regular) CW-complex structure ?
  • Can we extend this definition for all the topological spaces ?
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  • $\begingroup$ Yeah... I actually hate CW complexes ; in Hatcher's book I feel like it is the most handwavy explanation I ever read. Maybe I am just not a CW complex kind of guy, but this thing is seriously vague to me. I don't understand most of it, nor do I understand your answer... sorry. (It's not your fault.) $\endgroup$ Sep 2 '13 at 2:39
  • $\begingroup$ @PatrickDaSilva : perhaps a good way to understand what does it mean, is by taking an example : look at various simplicial complexes structure (instead of CW complex) on a disk, and see that we obtain its boundary by this process. I hope I have helped you, else, never mind ! $\endgroup$ Sep 2 '13 at 6:37
  • $\begingroup$ Yeah no, didn't help. I think I just need to find various points of view until one of them rings a bell and then I'll figure it out. But thanks for trying, I appreciate it. $\endgroup$ Sep 2 '13 at 16:07
  • $\begingroup$ @PatrickDaSilva : I have edited an explicit example. Maybe it's more understandable now. $\endgroup$ Sep 2 '13 at 18:27
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    $\begingroup$ @PatrickDaSilva : I warn you that if the local dimension of the space is constant, then we can probably only deal with n-cells (with n the global dimension), as you write and as my example; but if the local dimension is not constant then we need to deal with r-cells with r≤n. I don't know if this definition depends on the choice, that why I ask, but I guess no (with an easy proof). $\endgroup$ Sep 3 '13 at 19:49

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