A description/proof of that induced map from domain quotient of continuous map is continuous
Topics
About: topology space
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of quotient of set.
- The reader knows a definition of quotient map.
- The reader knows a definition of quotient topology.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map, its induced map (if exists) from any domain quotient is continuous.
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 topological spaces, \(M_1\) and \(M_2\), any continuous map, \(f: M_1 \rightarrow M_2\), and any quotient of \(M_1\), \(M_1/\sim\), and the quotient map, \(\rho: M_1 \rightarrow M_1/\sim\), if f equals for all the elements of \(M_1\) that correspond to each element of \(M_1/\sim\), there is the induced map, \(\tilde{f}\), as in \(f: M_1 \xrightarrow [\rho]{} M_1/\sim \xrightarrow[\tilde{f}]{} M_2\), and \(\tilde{f}\) is continuous.
2: Proof
For any open set, \(U \subseteq M_2\), \(f^{-1} (U)\) is open. \(f^{-1} (U) = \rho^{-1} \circ \tilde{f}^{-1} (U)\), because for any \(p \in f^{-1} (U)\), \(\tilde{f} \circ \rho (p) \in U\), \(\rho (p) \in \tilde{f}^{-1} (U)\), \(p \in \rho^{-1} \circ \tilde{f}^{-1} (U)\); on the other hand, for any \(p \in \rho^{-1} \circ \tilde{f}^{-1} (U)\), \(\rho (p) \in \tilde{f}^{-1} (U)\), \(\tilde{f} \circ \rho (p) \in U\) where \(\tilde{f} \circ \rho (p) = f (p)\), so, \(p \in f^{-1} (U)\). By the definition of quotient topology, if \(\rho^{-1} \circ \tilde{f}^{-1} (U)\) is open (in fact, it is), \(\tilde{f}^{-1} (U)\) is open, so, as for any open set, \(U \subseteq M_2\), \(\tilde{f}^{-1} (U)\) is open, \(\tilde{f}^{-1}\) is continuous.
