A description/proof of that topological space is connected if quotient space and each element of quotient space are connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quotient topology on set with respect to map.
- The reader knows a definition of connected topological space.
- The reader knows a definition of subspace topology.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is connected if a quotient space of it and each element of the quotient space are connected.
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
Any topological space, \(T\), is connected if a quotient space, \(T/\pi\), where \(\pi: T \rightarrow T/\pi\), and each element, \(p \in T/\pi\), as the subspace of \(T\) are connected.
2: Proof
Let us suppose that \(T/\pi\) and each \(p\) are connected. Let us suppose that \(T\) was not connected. There would be nonempty open sets, \(U_1, U_2 \subseteq T\), such that \(T = U_1 \cup U_2\) and \(U_1 \cap U_2 = \emptyset\). \(U_i \cap p\) is open on \(p\) by the definition of subspace topology. \(p = (U_1 \cap p) \cup (U_2 \cap p)\), \((U_1 \cap p) \cap (U_2 \cap p) = \emptyset\). As \(p\) is connected, \(U_1 \cap p = \emptyset\) or \(U_2 \cap p = \emptyset\). So, \(U_i = \cup_{\alpha \in A_i} p_\alpha\) where \(A_1 \cup A_2\) is an indexes set for the elements of \(T/\pi\) and \(A_1 \cap A_2 = \emptyset\). \(U_i = \pi^{-1} (\{p_\alpha\vert \alpha \in A_i\})\), and \(\{p_\alpha\vert \alpha \in A_i\}\) is open on \(T/\pi\) by the definition of quotient topology. \(T/\pi = \{p_\alpha \in T/\pi\vert \alpha \in A_1\} \cup \{p_\alpha \in T/\pi\vert \alpha \in A_2\}\) and \(\{p_\alpha \in T/\pi\vert \alpha \in A_1\} \cap \{p_\alpha \in T/\pi\vert \alpha \in A_2\} = \emptyset\) while \(\{p_\alpha \in T/\pi\vert \alpha \in A_i\} \neq \emptyset\), a contradiction against \(T/\pi\)'s being connected.
