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?



Manifolds with two coordinate charts
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**](http://dx.doi.org/10.2307/2034881), Proc. Am. Math. Soc. 12, 812-814 (1961). [ZBL0103.39305](https://zbmath.org/?q=an:0103.39305), [MR0126835 (23 #A4129)](https://www.ams.org/mathscinet-getitem?mr=126835).


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 $\psi(K) \subset U\_2$.




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**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.




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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**](http://dx.doi.org/10.1090/S0002-9904-1960-10400-4), Bull. Am. Math. Soc. 66, 74-76 (1960). [ZBL0132.20002](https://zbmath.org/?q=an:0132.20002), [MR0117695 (22 #8470b)](https://www.ams.org/mathscinet-getitem?mr=117695).


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**](http://dx.doi.org/10.2307/1970177), Ann. Math. (2) 75, 331-341 (1962). [ZBL0201.56202](https://zbmath.org/?q=an:0201.56202), [MR0133812 (24 #A3637)](https://www.ams.org/mathscinet-getitem?mr=133812).


that this implies that the sphere is bicollared. See


*Connelly, R.*, [**A new proof of Brown’s collaring theorem**](http://dx.doi.org/10.2307/2037284), Proc. Am. Math. Soc. 27, 180-182 (1971). [ZBL0208.50704](https://zbmath.org/?q=an:0208.50704), [MR0267588 (42 #2490)](https://www.ams.org/mathscinet-getitem?mr=267588).


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


