A description/proof of that restriction of continuous map on domain and codomain is continuous
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 continuous map.
- The reader knows a definition of subspace topology.
Target Context
- The reader will have a description and a proof of the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
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, \(T_1, T_2\), any continuous map, \(f: T_1 \rightarrow T_2\), any subset, \(S_1 \subseteq T_1\), and any subset, \(S_2 \subseteq T_2\) such that \(f (S_1) \subseteq S_2\), \(f|_{S_1}: S_1 \rightarrow S_2\) is continuous.
2: Proof
For any open set, \(U \subseteq S_2\), \(U = U' \cap S_2\) where \(U' \subseteq T_2\) is open on \(T_2\), by the definition of subspace topology. \({f|_{S_1}}^{-1} (U) = {f|_{S_1}}^{-1} (U' \cap S_2) = f^{-1} (U') \cap S_1\), because for any \(p \in {f|_{S_1}}^{-1} (U' \cap S_2)\), \(f|_{S_1} (p) \in U' \cap S_2\) by the definition of preimage, \(f (p) \in U' \cap S_2 \subseteq U'\), so, \(p \in f^{-1} (U')\), while of course \(p \in S_1\) because \(S_1\) is the domain of \({f|_{S_1}}^{-1}\); for any \(p \in f^{-1} (U') \cap S_1\), \(f|_{S_1} (p) \in U'\) because as \(p\) is on \(S_1\), \(f|_{S_1}\) can operate on it with the same result, which is on \(U'\), but as \(f|_{S_1} (S_1) = f (S_1) \subseteq S_2\), \(f|_{S_1} (p) \in U' \cap S_2\), so, \(p \in {f|_{S_1}}^{-1} (U' \cap S_2)\). As \(f\) is continuous, \(f^{-1} (U')\) is open on \(T_1\), and \(f^{-1} (U') \cap S_1\) is open on \(S_1\) by the definition of subspace topology.
