A description/proof of that identity map with domain and codomain having different topologies is continuous iff domain is finer than codomain
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of finer and coarser topologies.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a description and a proof of the proposition that any identity map with domain and codomain having the same set but having different topologies is continuous if and only if the domain topology is finer than codomain's.
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, \(T_1, T_2\), that have the same set but have different topologies and the identity map, \(f: T_1 \rightarrow T_2\), \(f\) is continuous if and only if the \(T_1\) topology is finer than the \(T_2\) topology.
2: Proof
Suppose that the \(T_1\) topology is finer than the \(T_2\) topology. Take any open set, \(U \subseteq T_2\). The preimage \(f^{-1} (U) = U\), is open on \(T_1\), because the \(T_1\) topology is finer, so, \(f\) is continuous.
Suppose that \(f\) is continuous. Take any open set, \(U \subseteq T_2\). The preimage \(f^{-1} (U) = U\), is open on \(T_1\), because \(f\) is continuous, so, the \(T_1\) topology is finer.
