description/proof of that connected topological manifold is path-connected
Topics
About: topological manifold
The table of contents of this article
Starting Context
- The reader knows a definition of topological manifold.
- The reader knows a definition of connected topological space.
- The reader knows a definition of path-connected topological space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that any connected topological manifold is path-connected.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
Any connected topological manifold, M, is path-connected.
2: Proof
For any point, \(p \in M\), take the set, S, as all the points path-connected with p, which is an open set, because for any \(p_1 \in S\), there is a neighborhood, \(U_{p_1}\), and a homeomorphism, \(\phi_1: U_{p_1} \rightarrow U_{\phi_1 (p_1)}\) where \(U_{\phi_1 (p_1)}\) is an open set on \(\mathbb {R}^n\), but any point, \(p_2 \in U_{p_1}\), is path-connected with p, because \(p_2\) is path-connected with \(p_1\) (\(\phi_1 (p_2)\) is path-connected with \(\phi_1 (p_1)\), and the path can be mapped into \(U_{p_1}\) by the continuous \(\phi_1^{-1}\)), but \(p_2\) can be path-connected with p via \(p_1\), so, \(U_{p_1} \subseteq S\), so, by the local criterion for openness, S is an open set; also \(M \setminus S\) is an open set, because for any \(p_3 \in M \setminus S\), there is a neighborhood, \(U_{p_3}\), and a homeomorphism, \(\phi_2: U_{p_3} \rightarrow U_{\phi (p_3)}\) where \(U_{\phi_2 (p_3)}\) is an open set on \(\mathbb {R}^n\), but any point, \(p_4 \in U_{p_3}\), is not path-connected with p, because \(p_4\) is path-connected with \(p_3\) (\(\phi_2 (p_4)\) is path-connected with \(\phi_2 (p_3)\), and the path can be mapped into \(U_{p_3}\) by the continuous \(\phi_2^{-1}\)), but if \(p_4\) was path-connected with p, also \(p_3\) would be path-connected with p via \(p_4\), a contradiction, so, \(U_{p_3} \subseteq M \setminus S\), so, by the local criterion for openness, \(M \setminus S\) is an open set; now S and \(M \setminus S\) are disjoint, \(M = S \cup (M \setminus S)\), and M is connected, so, one of the 2 subsets has to be empty, but S is not empty containing at least p, so, \(M \setminus S\) is empty.
3: Note
Note that the proposition is not about topological space in general, but about topological manifold.
