A description/proof of that map from mapping cylinder into topological space is continuous iff induced maps from adjunction attaching origin space and from adjunction attaching destination space are continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of mapping cylinder.
Target Context
- The reader will have a description and a proof of the proposition that for any mapping cylinder, any map from the mapping cylinder into any topological space is continuous if and only if the induced map from the adjunction attaching origin space and the induced map from the adjunction attaching destination space are 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, \(T_1, T_2, T_3\), any continuous map, \(f: T_1 \to T_2\), the mapping cylinder, \(M_f := T_2 \cup_{f_0} (T_1 \times I)\) where \(f_0: T_1 \times \{0\} \to T_2, \langle p, 0 \rangle \mapsto f (p)\) and \(I := [0, 1]\), any map, \(f_1: M_f \to T_3\), the induced map, \(f_2: T_1 \times I \to M_f \to T_3\), and the induced map, \(f_3: T_2 \to M_f \to T_3\), \(f_1\) is continuous if and only if \(f_2\) and \(f_3\) are continuous.
2: Proof
There is the quotient map, \(q: T_1 \times I + T_2 \to M_f\), where any \(p \in f_0 (T_1 \times \{0\})\) and \({f_0}^{-1} (p)\) are identified. \(f_2 = f_1 \circ q\vert_{T_1 \times I}\) and \(f_3 = f_1 \circ q\vert_{T_2}\) by definitions.
Let us prove that for any subset, \(S \subseteq T_3\), \(q^{-1} ({f_1}^{-1} (S)) = {f_2}^{-1} (S) + {f_3}^{-1} (S)\). For any \(p \in q^{-1} ({f_1}^{-1} (S))\), \(p \in T_1 \times I\) or \(p \in T_2\). If \(p \in T_1 \times I\), \(f_2 (p) \in S\), because \(q (p) \in {f_1}^{-1} (S)\) and \(f_1 \circ q (p) \in S\), but \(f_2 = f_1 \circ q\vert_{T_1 \times I}\). If \(p \in T_2\), \(f_3 (p) \in S\), because \(f_1 \circ q (p) \in S\), but \(f_3 = f_1 \circ q\vert_{T_2}\). So, \(p \in {f_2}^{-1} (S) + {f_3}^{-1} (S)\). For any \(p \in {f_2}^{-1} (S) + {f_3}^{-1} (S)\), \(f_2 (p) \in S\) or \(f_3 (p) \in S\), which means that \(f_1 \circ q\vert_{T_1 \times I} (p) \in S\) or \(f_1 \circ q\vert_{T_2} (p) \in S\). So, \(f_1 \circ q (p) \in S\), so, \(p \in q^{-1} ({f_1}^{-1} (S))\).
Let us suppose that \(f_2\) and \(f_3\) are continuous. For any open subset, \(U \subseteq T_3\), is \({f_1}^{-1} (U)\) open on \(M_f\)? It is about whether \(q^{-1} ({f_1}^{-1} (U))\) is open on \(T_1 \times I + T_2\), by the definition of quotient topology. \(q^{-1} ({f_1}^{-1} (U)) = {f_2}^{-1} (U) + {f_3}^{-1} (U)\), as is proven above, and it is open on \(T_1 \times I + T_2\).
Let us suppose that \(f_1\) is continuous. For any \(U \in T_3\), \(q^{-1} ({f_1}^{-1} (U)) = {f_2}^{-1} (U) + {f_3}^{-1} (U)\) is open on \(T_1 \times I + T_2\). \({f_2}^{-1} (U)\) is open on \(T_1 \times I\) and \({f_3}^{-1} (U)\) is open on \(T_2\).
