A description/proof of that quotient topology is sole finest topology that makes map continuous
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 quotient topology on set with respect to map.
- The reader knows a definition of map.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any set, and any surjection from the topological space to the set, the quotient topology is the sole finest topology that makes the map 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 space, \(T\), any set, \(S\), and any map, \(f: T \rightarrow S\), the quotient topology on \(S\) with respect to \(f\), \(O\), is the finest topology that makes \(f\) continuous, which means not only that \(O\) is a finest topology but also that any topology that makes \(f\) continuous is coarser than \(O\).
2: Proof
Any topology, \(O'\), that does not equal \(O\) but is not coarser than \(O\) has an open set, \(U\), that is not in \(O\). By the definition of \(O\), \(f^{-1} (U)\) is not open. So, \(f\) is not continuous.
