A description/proof of that subset of quotient topological space is closed iff preimage of subset under quotient map is closed
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 quotient map.
- The reader knows a definition of closed set.
- The reader admits the proposition that the preimage of the codomain minus any codomain subset of any map is the domain minus the preimage of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any quotient topological space, any subset is closed if and only if the preimage of the subset under the quotient map is closed.
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, \(T_1\), and any quotient topological space, \(T_2\), with respect to any quotient map, \(f: T_1 \rightarrow T_2\), any subset, \(C \subseteq T_2\), is closed if and only if \(f^{-1} (C)\) is closed.
2: Proof
Suppose that \(f^{-1} (C)\) is closed. \(f^{-1} (T_2 \setminus C) = T_1 \setminus f^{-1} (C)\) by the proposition that the preimage of the codomain minus any codomain subset of any map is the domain minus the preimage of the subset. \(T_1 \setminus f^{-1} (C)\) is open, so, \(T_2 \setminus C\) is open by the definition of quotient map, so, \(C\) is closed.
Suppose that \(C\) is closed. \(T_2 \setminus C\) is open. \(f^{-1} (T_2 \setminus C) = T_1 \setminus f^{-1} (C)\) as before, and is open as the preimage of an open set under a continuous map, so, \(f^{-1} (C)\) is closed.
