2023-03-12

234: For Injective Closed Map Between Topological Spaces, Inverse of Codomain-Restricted-to-Range Map Is Continuous

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that for injective closed map between topological spaces, inverse of codomain-restricted-to-range map is continuous

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any injective closed map between any topological spaces, the inverse of the codomain-restricted-to-range map 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, \(T_1\) and \(T_2\), any injective closed map, \(f: T_1 \rightarrow T_2\), and \(f\)'s restriction on the codomain, \(f': T_1 \rightarrow f (T_1)\), \(f'^{-1}: f (T_1) \rightarrow T_1\) is continuous.


2: Proof


For any closed set, \(C \subseteq T_1\), is \({f'^{-1}}^{-1} (C)\) closed on \(f (T_1)\)? As \(f'\) is a bijection, \({f'^{-1}}^{-1} (C) = f' (C)\), by the proposition that for any bijection, the preimage of any subset under the inverse of the map is the image of the subset under the map. As \(f\) is closed, \(f (C) = f' (C)\) is closed on \(T_2\), and \(f' (C) = f' (C) \cap f (T_1)\) is closed on \(f (T_1)\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset. By the proposition that if the preimage of any closed set under a topological spaces map is closed, the map is continuous, \(f'^{-1}\) is continuous.


References


<The previous article in this series | The table of contents of this series | The next article in this series>