description/proof of that adjunction topological space is Hausdorff if attaching-destination space is Hausdorff, attaching-origin space is regular, and domain of attaching-map is closed and retract of open neighborhood
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- 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 Hausdorff topological space.
- The reader knows a definition of regular topological space.
- The reader knows a definition of closed set.
- The reader knows a definition of retract of topological space.
- The reader knows a definition of neighborhood of subset.
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any adjunction topological space is Hausdorff if the attaching-destination space is Hausdorff, the attaching-origin space is regular, and the domain of the attaching-map is closed and a retract of an open neighborhood of the domain.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the regular topological spaces }\}\)
\(T_2\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
\(S\): \(\in \{\text{ the closed subsets of } T_1\}\)
\(f\): \(S \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(T_2 \cup_f T_1\): \(= \text{ the adjunction topological space }\)
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Statements:
\(\exists U_S \subseteq T_1 \in \{\text{ the open neighborhoods of } S\}, \exists r: U_S \to S \in \{\text{ the retractions }\}\)
\(\implies\)
\(T_2 \cup_f T_1 \in \{\text{ the Hausdorff topological spaces }\}\)
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2: Natural Language Description
For any regular topological space, \(T_1\), any Hausdorff topological space, \(T_2\), any closed subset, \(S \subseteq T_1\), that is a retract of any open neighborhood of \(S\), \(U_S \subseteq T_1\), where \(r: U_S \to S\) is the retraction, and any continuous map, \(f: S \to T_2\), the adjunction topological space, \(T_2 \cup_f T_1\), is Hausdorff.
3: Proof
Whole Strategy: Step 1: adopt the rule that we take the representative \(p\) of \([p]\) as \(p \in T_2\) or \(p \in T_1 \setminus S\), which makes the representative unique for each \([p]\); Step 2: for any \([p_1], [p_2] \in T_2 \cup_f T_1\) such that \([p_1] \neq [p_2]\), deal with the case that \(p_1, p_2 \in T_2\); Step 3: deal with the case that \(p_1, p_2 \in T_1 \setminus S\); Step 4: deal with the case that \(p_1 \in T_1 \setminus S\) and \(p_2 \in T_2\).
Step 1:
Let us denote the canonical maps as \(f_1: S \to T_1\), \(f_2: T_1 \to T_2 \cup_f T_1\), \(f_3: T_2 \to T_2 \cup_f T_1\), and \(f_4: T_1 + T_2 \to T_2 \cup_f T_1\).
Let us adopt the rule that we take the representative \(p\) of \([p]\) as \(p \in T_2\) or \(p \in T_1 \setminus S\), which makes the representative unique for each \([p]\).
It is about taking some open neighborhoods around any \([p_1], [p_2] \in T_2 \cup_f T_1\) such that \([p_1] \neq [p_2]\), \(U_{[p_1]}, U_{[p_2]} \subseteq T_2 \cup_f T_1\), such that \(U_{[p_1]} \cap U_{[p_2]} = \emptyset\).
Step 2:
Let us suppose that \(p_1, p_2 \in T_2\).
\(p_1 \neq p_2\). There are some open neighborhoods, \(U_{p_1}, U_{p_2} \subseteq T_2\), such that \(U_{p_1} \cap U_{p_2} = \emptyset\), because \(T_2\) is Hausdorff. \(f \circ r: U_S \to T_2\) is continuous. \((f \circ r)^{-1} (U_{p_i})\) is open on \(U_S\), so, is open on \(T_1\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space. \((f \circ r)^{-1} (U_{p_1})\) and \((f \circ r)^{-1} (U_{p_2})\) are disjoint, by the proposition that the preimages of any disjoint subsets under any map are disjoint.
Let us define \(U'_i := f_3 (U_{p_i}) \cup f_2 ((f \circ r)^{-1} (U_{p_i}))\). \(U'_1\) and \(U'_2\) are disjoint, because for any \([p] \in U'_i\), \([p] \in f_3 (U_{p_i})\) or \([p] \in f_2 ((f \circ r)^{-1} (U_{p_i}))\), \(p \in U_{p_i}\) or \(p \in (f \circ r)^{-1} (U_{p_i}) \cap (U_S \setminus S)\), because while \((f \circ r)^{-1} (U_{p_i}) = ((f \circ r)^{-1} (U_{p_i}) \cap S) \cup ((f \circ r)^{-1} (U_{p_i}) \cap (U_S \setminus S))\), \(f ((f \circ r)^{-1} (U_{p_i}) \cap S) \subseteq U_{p_i}\), because while \((f \circ r)^{-1} = r^{-1} \circ f^{-1}\), for each \(p' \in (r^{-1} \circ f^{-1} (U_{p_i})) \cap S\), \(p' \in f^{-1} (U_{p_i})\), because \(r\) does not move any point on \(S\). \(U'_i\) is open on \(T_2 \cup_f T_1\), because \({f_4}^{-1} (U'_i) \cap T_1 = ((f \circ r)^{-1} (U_i) \cap (U_S \setminus S)) \cup f^{-1} (U_i) = (f \circ r)^{-1} (U_i)\), open on \(T_1\), and \({f_4}^{-1} (U'_i) \cap T_2 = U_i\), open on \(T_2\).
Step 3:
Let us suppose that \(p_1, p_2 \in T_1 \setminus S\).
As \(T_1\) is regular, \(T_1\) is Hausdorff, and there are some open neighborhoods, \(U_{p_1}, U_{p_2} \subseteq T_1\), such that \(U_{p_1} \cap U_{p_2} = \emptyset\). As \(S\) is closed, \(T_1 \setminus S\) is open on \(T_1\), \(U'_i := U_{p_i} \cap (T_1 \setminus S)\) is open on \(T_1\), \(f_2 (U'_i)\) is open on \(T_2 \cup_f T_1\), \(f_2 (U'_1) \cap f_2 (U'_2) = \emptyset\).
Step 4:
Let us suppose that \(p_1 \in T_1 \setminus S\) and \(p_2 \in T_2\).
As \(T_1\) is regular and \(S\) is closed, there are some open neighborhoods of \(p_1\) and \(S\), \(U_{p_1} \subseteq T_1\) and \(U'_S \subseteq T_1\), such that \(U_{p_1} \cap U'_S = \emptyset\). Let us take the restriction of \(r\), \(r' := r\vert_{U_S \cap U'_S}: U_S \cap U'_S \to S\), continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. There is an open neighborhood of \(p_2\), \(U_{p_2} \subseteq T_2\). \(f \circ r'\) is continuous, \((f \circ r')^{-1} (U_{p_2})\) is open on \(U_S \cap U'_S\), and so is on \(T_1\) as \(U_S \cap U'_S\) is open on \(T_1\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Let us define \(U'_1 := f_2 (U_{p_1})\) and \(U'_2 := f_3 (U_{p_2}) \cup f_2 ((f \circ r')^{-1} (U_{p_2}))\). \(U'_1\) and \(U'_2\) are disjoint, because for any \([p] \in U'_2\), \([p] \in f_3 (U_{p_2})\) or \([p] \in f_2 ((f \circ r')^{-1} (U_{p_2}))\), \(p \in U_{p_2}\) or \(p \in (f \circ r')^{-1} (U_{p_2}) \cap ((U_S \cap U'_S) \setminus S)\), as before. \(U'_1\) is open on \(T_2 \cup_f T_1\), because \({f_4}^{-1} (U'_1) \cap T_1 = U_{p_1}\), open on \(T_1\), and \({f_4}^{-1} (U'_1) \cap T_2 = \emptyset\), open on \(T_2\). \(U'_2\) is open on \(T_2 \cup_f T_1\), because \({f_4}^{-1} (U'_2) \cap T_1 = ((f \circ r')^{-1} (U_{p_2}) \cap (U'_S \setminus S)) \cup f^{-1} (U_{p_2}) = (f \circ r')^{-1} (U_{p_2})\), open on \(T_1\), and \({f_4}^{-1} (U'_{p_2}) \cap T_2 = U_{p_2}\), open on \(T_2\).
