A description/proof of that for disjoint union topological space, inclusion from constituent topological space to disjoint topological space is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of disjoint union topology.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a description and a proof of the proposition that for any disjoint union topological space, the inclusion from any constituent topological space to the disjoint topological space 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 the set of any topological spaces, \(\{T_\alpha\}\), and the disjoint union topological space, \(T := \coprod_\alpha T_\alpha\), each inclusion, \(f_\alpha: T_\alpha \rightarrow T\), is continuous.
2: Proof
For any open set, \(U \subseteq T\), \(U_\alpha := U \cap T_\alpha\) is open on \(T_\alpha\) by the definition of disjoint union topology. \((f_\alpha)^{-1} (U) = U_\alpha\), because for any \(p \in (f_\alpha)^{-1} (U)\), \(p \in T_\alpha\) and \(p \in U\), so, \(p \in U \cap T_\alpha\); for any \(p \in U_\alpha\), \(p \in T_\alpha\) and \(p \in U\), \(f_\alpha (p) \in U\), so, \(p \in (f_\alpha)^{-1} (U)\). So, \((f_\alpha)^{-1} (U)\) is open on \(T_\alpha\).
