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?

  • $\begingroup$ 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. $\endgroup$ – Misha Dec 28 '12 at 23:38
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    $\begingroup$ The second one follows from the first and the Schoenflies theorem by doubling. $\endgroup$ – Ian Agol Dec 28 '12 at 23:52
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    $\begingroup$ @Agol: do you mean the Brown-Mazur theorem, i.e. the collared version of the Schoenflies theorem? $\endgroup$ – Lee Mosher Dec 29 '12 at 0:03
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    $\begingroup$ 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? $\endgroup$ – Ian Agol Dec 29 '12 at 0:53
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    $\begingroup$ 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. $\endgroup$ – Siqi He Dec 29 '12 at 9:09

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.

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  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Dec 29 '12 at 13:32
  • $\begingroup$ @Lee Mosher : Something like that works! I added it in an edit. $\endgroup$ – Andy Putman Dec 29 '12 at 16:59
  • $\begingroup$ Very nice. I would accept Ian Agol's comment too if I could. $\endgroup$ – Lee Mosher Dec 30 '12 at 14:59
  • $\begingroup$ 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. $\endgroup$ – Ricardo Andrade May 19 '15 at 23:49

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