# Manifolds with two coordinate charts

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?

• 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. Dec 28, 2012 at 23:38
• The second one follows from the first and the Schoenflies theorem by doubling. Dec 28, 2012 at 23:52
• @Agol: do you mean the Brown-Mazur theorem, i.e. the collared version of the Schoenflies theorem? Dec 29, 2012 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? Dec 29, 2012 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. Dec 29, 2012 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

Brown, Morton, The monotone union of open (n)-cells is an open (n)-cell, Proc. Am. Math. Soc. 12, 812-814 (1961). ZBL0103.39305, MR0126835 (23 #A4129).

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

Brown, Morton, A proof of the generalized Schoenflies theorem, Bull. Am. Math. Soc. 66, 74-76 (1960). ZBL0132.20002, MR0117695 (22 #8470b).

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

Brown, Morton, Locally flat imbeddings of topological manifolds, Ann. Math. (2) 75, 331-341 (1962). ZBL0201.56202, MR0133812 (24 #A3637).

that this implies that the sphere is bicollared. See

Connelly, R., A new proof of Brown’s collaring theorem, Proc. Am. Math. Soc. 27, 180-182 (1971). ZBL0208.50704, MR0267588 (42 #2490).

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

• 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. Dec 29, 2012 at 13:32
• @Lee Mosher : Something like that works! I added it in an edit. Dec 29, 2012 at 16:59
• Very nice. I would accept Ian Agol's comment too if I could. Dec 30, 2012 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. May 19, 2015 at 23:49