I'm reading Lee, 'Introduction to Topological Manifolds', 2011. After he introduces $n$-manifold with a boundary

An $n$-manifold with a boundary is a second countable Hausdorff space in which any point has a neighborhood which is homeomorphic either to an open subset of $\mathbb R^n$ or to an open subset of $\mathbb H^n = \{x\in \mathbb R^n:x_n\geq 0\}$ endowed with a Euclidean topology.

The interior chart is then $(U,\phi)$ where $U$ is open in $M$ and $\phi:U\to V\subseteq \mathbb R^n$ is a homeomorphism (here $V$ is open in $\mathbb R^n$) and boundary chart as $(U',\phi')$ where $U'$ is open in $M$ and $\phi':U'\to V'\subseteq \mathbb H^n$ is a homeomorphism (here $V$ is open in $\mathbb H^n$ and $V'\cap \partial \mathbb H^n\neq\emptyset$).

Then he writes:

If M is an $n$-manifold with boundary, a point $p$ in $M$ is called an interior point of $M$ if it is in the domain of an interior chart; and it is called a boundary point of $M$ if it is in the domain of a boundary chart that takes $p$ to $\partial \mathbb H^n$.

The natural question is - can be $p$ both interior and boundary point of $M$? In 2011 version of the book Lee writes:

Every point of M is either an interior point or a boundary point: if $p \in M$ is in the domain of an interior chart, then it is an interior point; on the other hand, if it is in the domain of a boundary chart, then it is an interior point if its image lies in $\operatorname{Int}\mathbb H^n$, and a boundary point if the image lies in $\partial \mathbb H^n$.

I cannot understand his argumentation. Moreover, in the very same book on the next page he writes explicitly that it is not possible to prove that $p$ is either interior or boundary at that stage. (Theorem 2.59). I have two question:

  1. Do I miss something, or he states two contradictory facts?

  2. I tried to prove that any point is either boundary interior or boundary and come to the fact that any subset $V'$ open in $\mathbb H^n$ such that $V'\cap \partial \mathbb H^n\neq\emptyset$ is not homeomorphic to any open subset of $\mathbb R^n$ and couldn't prove that fact. Can you help me with that proof?

  • $\begingroup$ ad 2: Let $V \subset \mathbb H^n$ open and $x \in V \cap \partial \mathbb H^n$. Suppose $V$ is homeomorphic to an open subset of $\mathbb R^n$, then clearly there is some neighborhood $U$ of $x$ homeomorphic to the open ball $B^n$. Now $U \setminus \{ x \}$ is homotopic to $U$ but $B^n \setminus \{ \ast \}$ is homotopic to $S^{n-1}$. $\endgroup$ Oct 19, 2011 at 17:27
  • $\begingroup$ He is not stating contradictory facts. Notice if a point is in the interior of boundary chart, there exists an open ball about the image of that point. Cut down the chart so we map just to this open ball and we'll have an interior chart. Thus the point is an interior point in the original sense as well. On the other hand if a point's image lies on the boundary of a boundary chart, you can't cut the chart down to avoid the boundary and so that point can't be an interior point -- unless one had a homeomorphism from a half ball to a ball which Alexander showed is impossible. $\endgroup$
    – Bill Cook
    Oct 19, 2011 at 21:55
  • $\begingroup$ @BillCook: Thanks, but I have the same way of proof I and wrote where I stacked. By contradiction I mean that his argument starting with: If $M$ is an $n$- manifold does not seem to be correct unless he states there that $p$ belongs to at least one of the classes. I read either, or as exactly one class. $\endgroup$
    – SBF
    Oct 19, 2011 at 22:03
  • $\begingroup$ Without knowing the book, my feeling is that he might not be able to prove the exclusive or at that point; his argument is clearly designed to show the non-exclusive or. You need the (nontrivial) theorem that states that the image of an injective continuous map from an open subset of $\mathbb{R}^n$ to $\mathbb{R}^n$ is open ("invariance of domain"). $\endgroup$ Oct 21, 2011 at 16:10

2 Answers 2


I think your confusion is based on interpreting "either ... or" as meaning "one or the other but not both" (exclusive or). Although this interpretation is not uncommon in everyday English, that's not how I was using the phrase in the passage you quoted. I meant "or" in the usual mathematical-logic sense of "one or the other or both" (inclusive or).

So my claim on page 43 is that "every point of $M$ is either an interior point or a boundary point," but I'm not claiming (at this stage) that a point cannot be both. The fact that it cannot be both is what Theorem 2.59 (Invariance of the Boundary) guarantees, but the tools for proving it in full generality are not developed until Chapter 13 (Homology).

The issue of exclusive vs. inclusive or is a common source of confusion. In ordinary English, "either...or" is probably most often interpreted as exclusive or, but not universally. The meaning depends on context: in "you can have either soup or salad with your entree," it's clearly exclusive, but in "you must have either a bachelor's degree or three years' experience," it's just as clearly inclusive. The same problem arises even more with "or" alone (if you delete "either" from those two sentences, it doesn't change the meaning of either one).

While there may be mathematicians who use "either A or B" when they really mean "either A or B but not both," I think most careful mathematical writers insert some phrase such as "but not both" when they really mean exclusive or, and otherwise interpret "or" and "either...or" as inclusive. That's certainly the way I write. I'm sorry it confused you -- I'll try to be more sparing with my use of "either" from now on.

John M. (Jack) Lee (the author)

  • 1
    $\begingroup$ I found the following example elsewhere in this site: "Tea or coffee?" vs. "Sugar or milk?", which sounds more attractive despite saying the same thing. :-) // This post is a mild surprise for me since I am currently reading your book independently. It's nicely written. Thank you for the work! $\endgroup$
    – Srivatsan
    Jan 21, 2012 at 18:28
  • 1
    $\begingroup$ @Srivatsan -- you're welcome! And thanks for the compliment. $\endgroup$
    – Jack Lee
    Jan 21, 2012 at 20:21
  • $\begingroup$ As for Srivatsan, your post is a surprise for me :-) I'm not a native English speaker, and I was taught that 'either ... or' is used universally as the 'exclusive or' - so I guess that's the main reason of my confusion. Thank you very much for the answer and for your books: while I'm reading them and have a question, the answer usually comes in the next paragraph. It's so helpful when one tries to learn from the book, not from the lectures. $\endgroup$
    – SBF
    Jan 22, 2012 at 12:51

The invariance of the boundary can be seen easily from the Local Inversion Theorem of Calculus. Indeed, it is enough to see that a diffeomorphism $h$ (a change of charts) of a halfspace $\mathbb H^n$ onto another $\mathbb H^n$ maps $\{x_n\ge0\}$ into $\{x_n\ge0\}$. But suppose, say, $h(0)=(0,\dots,0,a)$ with $a>0$. Then the inverse $h^{-1}$ restricts to a differentiable mapping $W=\{x_n>0\}\to\mathbb R^n$ with non zero jacobian det. Thus, by local inversion, $h^{-1}$ maps some open ball $B\subset W$ around $h(0)$ onto an open nbhd $V$ of $a$ in $\mathbb R^n$. Such a $V$ cannot be included in $\mathbb H^n$, and we are done.

Just for the record, this uses local inversion, but the result is in fact more truly elementary: it can be deduced by a little jiggling with the notion of derivative. Although this is not easy to find in the literature...


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