MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is an early reference for the fact that if a compact, connected $n$-manifold $M$ is covered by two open sets homeomorphic to $\mathbb{R}^n$ then $M$ is homeomorphic to $S^n$?

And is it true that if $M$ is a compact, connected $n$-manifold with boundary, and if $M$ is covered by two open sets homeomorphic to $\lbrace(x_1,\ldots,x_n) \in \mathbb{R}^n | x_n \ge 0\rbrace$, then $M$ is a closed ball?

share|cite|improve this question
Lee: I do not see how one could prove this without the annulus theorem, so the earliest reference (say in dimension at least 5) would be Kirby and Siebenmann; in dimension 4 Freedman and Quinn; in dimension 3 and lower I am not sure whom to attribute this result to. – Misha Dec 28 '12 at 23:38
The second one follows from the first and the Schoenflies theorem by doubling. – Ian Agol Dec 28 '12 at 23:52
@Agol: do you mean the Brown-Mazur theorem, i.e. the collared version of the Schoenflies theorem? – Lee Mosher Dec 29 '12 at 0:03
Sorry, I meant Newman in 1966 who proved the topological Poincare conjecture (or Smale, depending on which category you're interested in). Clearly your manifold is a homotopy sphere; the question is whether it was identified to be a sphere earlier than the proofs of the Poincare conjecture? – Ian Agol Dec 29 '12 at 0:53
I think this problem has some relation to Ljusternik-Schnirelmann category number. Ljusternik-Schnirelmann category number is the minimal number of coordinate that can cover the manifold. In dimension two, if a manifold is not S^2 that the category number is 3. I know there is a theorem that the minimal critical point of a function over manifold is bigger than the category number. I don't know whether these two number is equal in other dimension. – Siqi He Dec 29 '12 at 9:09
up vote 19 down vote accepted

I'll only discuss the first question (EDIT : Actually, I address the second question at the end). As Agol pointed out in the comments, for $n \geq 5$ this is an easy consequence of Newman's 1966 proof of the Poincare conjecture in the topological category.

I don't know if it was explicitly stated earlier than this. However, it can easily be derived from the main result of the paper

MR0126835 (23 #A4129) Brown, Morton The monotone union of open n-cells is an open n-cell. Proc. Amer. Math. Soc. 12 1961 812–814.

In fact, this works in all dimensions (including $3$ and $4$).

Brown's theorem is as follows. Assume that $M$ is a topological $n$-manifold and that for all compact $K \subset M$, there exists some open set $U \subset M$ with $K \subset U$ and $U \cong \mathbb{R}^n$. Then $M \cong \mathbb{R}^n$. Brown's proof is clever, but completely elementary.

To get the desired result from this, assume that $X = U_1 \cup U_2$ with $U_i \cong \mathbb{R}^n$ and that $X$ is compact. Let $\phi : \mathbb{R}^n \rightarrow U_1$ be a homeomorphism. It is enough to prove that $X \setminus \{\phi(0)\} \cong \mathbb{R}^n$. We will do this with Brown's theorem. Consider a compact set $K \subset X \setminus \{\phi(0)\}$. To verify Brown's criteria, it is enough to construct a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ such that $\psi(K) \subset U_2$.

For $r>0$, let $B(r) \subset \mathbb{R}^n$ be the ball of radius $r$. The set $U_1 \setminus U_2$ is compact, so there exists some $R>0$ such that $U_1 \setminus \phi(B(R)) \subset U_2$. Also, there exists some $\epsilon > 0$ such that $K \cap \phi(B(\epsilon)) = \emptyset$. It is easy to construct a homeomorphism $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $f(B(\epsilon)) = B(2R)$ and $f(0)=0$ and $f|_{\mathbb{R}^n \setminus B(3R)} = \text{id}$. We can therefore define a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ by $\psi(p) = \phi \circ f \circ \phi^{-1}(p)$ for $p \in U_1 \setminus \{\phi(0)\}$ and $\psi(p) = p$ for $p \notin U_1$. Clearly $\psi(K) \subset U_2$.

EDIT : Lee suggested that this might be able to address his second question too. I thought a bit about it, and I believe that it can. The key is the following "relative" version of Brown's theorem, which can be proven exactly like Brown's theorem.

Theorem : Let $(M,N)$ be a pair consisting of a topological $n$-manifold $M$ and a closed submanifold $N \subset M$. Assume that for all compact $K \subset M$, there exists some open set $U \subset M$ such that $K \subset U$ and such that the pair $(U,U \cap N)$ is homeomorphic to the pair $(\mathbb{R}^n,\mathbb{R}^{n-1})$ (the second embedded in the standard way). Then $(M,N) \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.

To apply this, assume that $X$ is a compact manifold with boundary and that $X = U_1 \cup U_2$ with $(U_i,\partial U_i) \cong (\mathbb{R}^n_{\geq 0},\mathbb{R}^{n-1})$. Double $X$ to get a closed manifold $Y$, and let $Y' \subset Y$ be the image of the boundary of $X$. The open sets $U_i$ double to give an open cover $Y = V_1 \cup V_2$. Letting $V_i' = V_i \cap Y'$, we have $(V_i,V_i') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$. Let $(M,M')$ be the result of deleting the image of $0$ in $(V_1,V_1')$. It is enough to prove that $(M,M') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$, and this can be proven just like above.

Of course, Agol answered the second question first -- it follows from the topological Schonfleiss theorem applied to the double, which was proven by Brown in

MR0117695 (22 #8470b) Reviewed Brown, Morton A proof of the generalized Schoenflies theorem. Bull. Amer. Math. Soc. 66 1960 74–76. 54.00 (57.00)

Mazur had earlier proven a weaker result. This requires the sphere to be bicollared, but this holds. Indeed, from the assumptions the sphere is locally bicollared, and Brown proved in

MR0133812 (24 #A3637) Brown, Morton Locally flat imbeddings of topological manifolds. Ann. of Math. (2) 75 1962 331–341.

that this implies that the sphere is bicollared. See

MR0267588 (42 #2490) Connelly, Robert A new proof of Brown's collaring theorem. Proc. Amer. Math. Soc. 27 1971 180–182.

for a super-easy proof of Brown's collaring theorem.

share|cite|improve this answer
If Brown's clever proof works for a closed half-space in place of $\mathbb{R}^n$, then your proof would extend to answer my second question as well. – Lee Mosher Dec 29 '12 at 13:32
@Lee Mosher : Something like that works! I added it in an edit. – Andy Putman Dec 29 '12 at 16:59
Very nice. I would accept Ian Agol's comment too if I could. – Lee Mosher Dec 30 '12 at 14:59
For a different reference, the proof in this answer of the first statement appears in theorem 2.7 of Steve Ferry's geometric topology notes. – Ricardo Andrade May 19 '15 at 23:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.