A description/proof of that for quotient map, its restriction on open or closed saturated domain and on restricted image codomain is quotient map
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quotient map.
- The reader knows a definition of saturated subset.
- The reader knows a definition of closed set.
- The reader knows a definition of subspace topology.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any open set on any open topological subspace is open on the base space.
- The reader admits 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.
- The reader admits the proposition that any closed set on any closed topological subspace is closed on the base space.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any quotient map, its restriction on any open or closed saturated subset domain and on the restricted image codomain is a quotient map.
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 quotient map, \(f: T_1 \rightarrow T_2\), and any open or closed saturated subset with respect to \(f\), \(S \subseteq T_1\), the restriction, \(f\vert_S \rightarrow f (S)\), is a quotient map.
2: Proof
\(f\vert_S\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. \(f\vert_S\) is surjective.
For any subset, \(S' \subseteq f (S)\), \((f\vert_S)^{-1} (S') = f^{-1} (S') \cap S\), because for any point, \(p \in (f\vert_S)^{-1} (S')\), \(f\vert_S (p) = f (p) \in S'\) and \(p \in S\). For any point, \(p \in f^{-1} (S') \cap S\), \(f (p) = f\vert_S (p) \in S'\). \(S' \subseteq f (S)\), \(f^{-1} (S') \subseteq f^{-1} f (S) = S\), where the last equation is by the definition of saturated subset. So, \(f^{-1} (S') \cap S = f^{-1} (S')\).
Let us suppose that \(S\) is open. For any subset, \(S' \subseteq f (S)\), such that \((f\vert_S)^{-1} (S')\) is open on \(S\), is \(S'\) open on \(f (S)\)? \((f\vert_S)^{-1} (S') = f^{-1} (S')\) is open on \(S\), and is open on \(T_1\), by the proposition that any open set on any open topological subspace is open on the base space. So, \(S'\) is open on \(T_2\), because \(f\) is a quotient map. \(S' = S' \cap f (S)\) is open on \(f (S)\), by the definition of subspace topology.
Let us suppose that \(S\) is closed. For any subset, \(S' \subseteq f (S)\), such that \((f\vert_S)^{-1} (S')\) is closed on \(S\), is \(S'\) closed on \(f (S)\)? \((f\vert_S)^{-1} (S') = f^{-1} (S')\) is closed on \(S\), and is closed on \(T_1\), by the proposition that any closed set on any closed topological subspace is closed on the base space. So, \(S'\) is closed on \(T_2\), by 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. \(S' = S' \cap f (S)\) is closed on \(f (S)\), by the definition of subspace topology. By 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, \(f\vert_S\) is a quotient map.
