A description/proof of that for quotient map, codomain subset is closed if preimage of subset 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 map.
- 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 map, any codomain subset is closed if the preimage of the subset 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 spaces, \(T_1\) and \(T_2\), any quotient map, \(f: T_1 \rightarrow T_2\), and any subset, \(S \subseteq T_2\), \(S\) is closed if \(f^{-1} (S)\) is closed.
2: Proof
Suppose that \(f^{-1} (S)\) is closed. \(f^{-1} (T_2 \setminus S) = T_1 \setminus f^{-1} (S)\), 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. As \(T_1 \setminus f^{-1} (S)\) is open, by the definition of quotient map, \(T_2 \setminus S\) is open. So, \(S\) is closed.
