Manifolds with two coordinate charts - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:38:56Z http://mathoverflow.net/feeds/question/117457 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117457/manifolds-with-two-coordinate-charts Manifolds with two coordinate charts Lee Mosher 2012-12-28T23:19:26Z 2013-01-11T12:56:04Z <p>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 $S^n$?</p> <p>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?</p> http://mathoverflow.net/questions/117457/manifolds-with-two-coordinate-charts/117475#117475 Answer by Andy Putman for Manifolds with two coordinate charts Andy Putman 2012-12-29T04:55:50Z 2012-12-29T16:58:54Z <p>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.</p> <p>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</p> <p>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. </p> <p>In fact, this works in all dimensions (including $3$ and $4$). </p> <p>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.</p> <p>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 <code>$X \setminus \{\phi(0)\} \cong \mathbb{R}^n$</code>. We will do this with Brown's theorem. Consider a compact set <code>$K \subset X \setminus \{\phi(0)\}$</code>. To verify Brown's criteria, it is enough to construct a homeomorphism <code>$\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$</code> such that $\psi(K) \subset U_2$. </p> <p>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 <code>$\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$</code> by $\psi(p) = \phi \circ f \circ \phi^{-1}(p)$ for <code>$p \in U_1 \setminus \{\phi(0)\}$</code> and $\psi(p) = p$ for $p \notin U_1$. Clearly $\psi(K) \subset U_2$.</p> <hr> <p>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.</p> <p>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 $(M,N) \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.</p> <p>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.</p> <hr> <p>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</p> <p>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)</p> <p>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</p> <p>MR0133812 (24 #A3637) Brown, Morton Locally flat imbeddings of topological manifolds. Ann. of Math. (2) 75 1962 331–341. </p> <p>that this implies that the sphere is bicollared. See </p> <p>MR0267588 (42 #2490) Connelly, Robert A new proof of Brown's collaring theorem. Proc. Amer. Math. Soc. 27 1971 180–182. </p> <p>for a super-easy proof of Brown's collaring theorem.</p>