A description/proof of equivalence of map continuousness in topological sense and in norm sense for coordinates functions
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold.
- The reader knows a definition of continuous map.
- The reader knows a definition of continuous, normed spaces map.
Target Context
- The reader will have a description and a proof of the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.
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
For any \(C^\infty\) manifolds, \(M_1\) and \(M_2\), and any map, \(f: M_1 \rightarrow M_2\), \(f\) is continuous in the topological sense if and only if \(f\) is continuous in the norm sense for the coordinates functions.
2: Proof
Suppose that \(f\) is continuous in the norm sense for the coordinates functions. For any open set, \(U_2 \subseteq M_2\), around any point, \(m_2 \in U_2\), there is a chart, \((U_{m_2}, \phi_{m_2}), \phi_{m_2}: U_{m_2} \rightarrow \phi_{m_2} (U_{m_2}) \subseteq \mathbb{R}^{d_2}\) where \(U_{m_2} \subseteq U_2\). \(f^{-1} (m_2)\) may consist of multiple points, but around any point, \(m_1 \in f^{-1} (m_2)\), there is a chart, \((U_{m_1}, \phi_{m_1}), \phi_{m_1}: U_{m_1} \rightarrow \phi_{m_1} (U_{m_1}) \subseteq \mathbb{R}^{d_1}\), and by the continuousness in the norm sense, for any open ball, \(B_{\phi_2 (m_2)-\epsilon} \subseteq \phi_{m_2} (U_{m_2})\), there is an open ball, \(B_{\phi_1 (m_1)-\delta} \subseteq \phi_{m_1} (U_{m_1})\), such that \(\phi_{m_2} (f (\phi_{m_1}^{-1} (B_{\phi_1 (m_1)-\delta}))) \subseteq B_{\phi_2 (m_2)-\epsilon}\). But \(\phi_{m_1}^{-1} (B_{\phi_1 (m_1)-\delta})\), open on \(M_1\), is included in \(f^{-1} (U_2)\), because \(f (\phi_{m_1}^{-1} (B_{\phi_1 (m_1)-\delta})) \subseteq \phi_{m_2}^{-1} (B_{\phi_2 (m_2)-\epsilon}) \subseteq U_2\). As \(\{m_1\} = f^{-1} (U_2)\), there is an open set, \(\phi_{m_1}^{-1} (B_{\phi_1 (m_1)-\delta}) \subseteq f^{-1} (U_2)\), at any point, \(m_1 \in f^{-1} (U_2)\), so, by the local criterion for openness, \(f^{-1} (U_2)\) is open. So, as the preimage of any open set is open, \(f\) is continuous in the topological sense.
Suppose that \(f\) is continuous in the topological sense. For any point, \(m_1 \in M_1\) where \(f (m_1) = m_2\), there is a chart, \((U_{m_2}, \phi_{m_2}), \phi_{m_2}: U_{m_2} \rightarrow \phi_{m_2} (U_{m_2}) \subseteq \mathbb{R}^{d_2}\). There is an open ball, \(B_{\phi_{m_2-\epsilon}} \subseteq \phi_{m_2} (U_{m_2})\), and by the continuousness in the topological sense, \(f^{-1} (\phi_{m_2}^{-1} (B_{\phi_{m_2}-\epsilon}))\) is open including \(m_1\), so, there is a chart, \((U_{m_1}, \phi_{m_1}), \phi_{m_1}: U_{m_1} \rightarrow \phi_{m_1} (U_{m_1}) \subseteq \mathbb{R}^{d_1}\) where \(U_{m_1} \subseteq f^{-1} (\phi_{m_2}^{-1} (B_{\phi_{m_2}-\epsilon}))\), and an open ball, \(B_{\phi_{m_1}-\delta} \subseteq \phi_{m_1} (U_{m_1})\), which satisfies \(\phi_{m_2} (f ({\phi_{m_1}}^{-1} (B_{\phi_{m_1}-\delta}))) \subseteq B_{\phi_{m_2}-\epsilon}\), because \(B_{\phi_{m_1}-\delta}\) is included in \(\phi_{m_1} (U_{m_1})\) and \(U_{m_1}\) is included in \(f^{-1} (\phi_{m_2}^{-1} (B_{\phi_{m_2}-\epsilon}))\), which means that \(f ({\phi_{m_1}}^{-1} (B_{\phi_{m_1}-\delta}))\) is included in \({\phi_{m_2}}^{-1} (B_{\phi_{m_2}-\epsilon})\), which means that \(\phi_{m_2} (f ({\phi_{m_1}}^{-1} (B_{\phi_{m_1}-\delta})))\) is included in \(B_{\phi_{m_2}-\epsilon}\).
3: Note
While prevalently the continuousness of a map in topological sense is claimed by the continuousness of coordinates functions in norm sense, the argument is valid only because of this proposition or a like.
