A description/proof of that continuous surjection between topological spaces is quotient map if any codomain subset is closed if its preimage is closed
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of surjection.
- The reader knows a definition of closed set.
- The reader knows a definition of quotient map.
- The reader admits the proposition that the preimage of the codomain minus any codomain subset under 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 any continuous surjection between any topological spaces is a quotient map if any codomain subset is closed if its preimage 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, \(T_1, T_2\), and any continuous surjection, \(f: T_1 \rightarrow T_2\), \(f\) is a quotient map if any subset, \(S \subseteq T_2\), is closed if its preimage, \(f^{-1} (S)\), is closed.
2: Proof
Let us suppose that \(S\) is closed if \(f^{-1} (S)\) is closed. For any subset, \(S' \subseteq T_2\), if \(f^{-1} (S')\) is open, is \(S'\) open? \(T_1 \setminus f^{-1} (S')\) is closed. \(T_1 \setminus f^{-1} (S') = f^{-1} (T_2 \setminus S')\), by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset. \(T_2 \setminus S'\) is closed. \(S'\) is open.
3: Note
Being continuous is required as a precondition, because the continuousness is not guaranteed by just any codomain subset's being closed if the preimage is close.
