A description/proof of that for adjunction topological space, canonical map from attaching-destination space to adjunction space is continuous embedding
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 adjunction topological space obtained by attaching topological space via continuous map to topological space.
- The reader knows a definition of continuous embedding.
- The reader knows a definition of subspace topology.
- The reader knows a definition of quotient topology on set with respect to map.
- The reader knows a definition of topological sum.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any adjunction topological space, the canonical map from the attaching-destination space to the adjunction space is a continuous embedding.
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 subset, \(S \subseteq T_1\), any continuous map, \(f: S \rightarrow T_2\), and the adjunction topological space, \(T_2 \cup_f T_1\), the canonical map, \(f_3: T_2 \rightarrow T_2 \cup_f T_1\), is a continuous embedding.
2: Proof
\(f_3\) is obviously injective. \(f_3\) is continuous, because \(f_3: T_2 \rightarrow T_1 + T_2 \rightarrow T_2 \cup_f T_1\) is a composition of continuous maps (the former constituent of the composition is so because of the definition of topological sum; the latter is so because of the definition of quotient space). \({f_3}': T_2 \rightarrow f_3 (T_2)\) is bijective and continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. The remaining issue is whether \({{f_3}'}^{-1}\) is continuous. For any open set, \(U \subseteq T_2\), \(U = S' \cup S''\) where \(S' = U \cap f (T_1)\) and \(S'' = U \setminus S'\). Is \({{{f_3}'}^{-1}}^{-1} (U)\) open on \(f_3 (T_2)\)? \({{{f_3}'}^{-1}}^{-1} (U) = {f_3}' (U) = {f_3}' (S') \cup {f_3}' (S'')\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets. \({f_3}' (S') = \{[p]: p \in S', [p] = \{f^{-1} (p) \cup \{p\}\}\}\); \({f_3}' (S'') = S''\).
Whether \({{{f_3}'}^{-1}}^{-1} (U)\) is open on \(f_3 (T_2)\) is whether there is an open set, \(U' \subset T_2 \cup_f T_1\), such that \({{{f_3}'}^{-1}}^{-1} (U) = U' \cap f_3 (T_2)\), by the definition of subspace topology, so, let us define such a \(U'\). \(f^{-1} (U)\) is open on \(S\), so, there is an open set, \(U'' \subseteq T_1\), such that \(f^{-1} (U) = U'' \cap S\). \(U' := (U'' \setminus S) \cup {{{f_3}'}^{-1}}^{-1} (U)\). Let us prove that \(U'\) is open on \(T_2 \cup_f T_1\), which is whether \(f_4^{-1} (U')\) is open on \(T_1 + T_2\) where \(f_4: T_1 + T_2 \rightarrow T_2 \cup_f T_1\) is the quotient map, by the definition of quotient topology on set with respect to map, which is whether \(f_4^{-1} (U') \cap T_1\) and \(f_4^{-1} (U') \cap T_2\) are open, by the definition of topological sum. \(f_4^{-1} (U') \cap T_1 = U''\); \(f_4^{-1} (U') \cap T_2 = U\). So, \(U'\) is open on \(T_2 \cup_f T_1\). \({{{f_3}'}^{-1}}^{-1} (U) = U' \cap f_3 (T_2)\), because \(U'\) is \((U'' \setminus S) \cup {{{f_3}'}^{-1}}^{-1} (U)\) while \(f_3 (T_2) \cap (U'' \setminus S) = \emptyset\).
