A description/proof of that map between topological spaces is continuous iff preimage of each closed subset of codomain 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 closed set.
- 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 topological spaces map is continuous if and only if the preimage of any closed subset of the codomain 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 map, \(f: T_1 \rightarrow T_2\), \(f\) is continuous if and only if for each close subset, \(C \in T_2\), the preimage, \(f^{-1} (C)\), is closed on \(T_1\).
2: Proof
Let us suppose that \(f^{-1} (C)\) is closed on \(T_1\). For any open subset, \(U \subseteq T_2\), \(C\) can be taken to be \(C:= T_2 \setminus U\), closed on \(T_2\). \(f^{-1} (U) = f^{-1} (T_2 \setminus C) = T_1 \setminus f^{-1} (C)\), by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset, and as \(f^{-1} (C)\) is closed on \(T_1\), \(f^{-1} (U)\) is open on \(T_1\).
Let us suppose that \(f\) is continuous. \(U:= T_2 \setminus C\) is open on \(T_2\). \(f^{-1} (C) = f^{-1} (T_2 \setminus U) = T_1 \setminus f^{-1} (U)\), 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, which is closed on \(T_1\).
