A description/proof of that subset of subspace of adjunction topological space is open iff projections of preimage of subset are open with condition.
Topics
About: topological space
The table of contents of this article
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 subspace topology.
- The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
- The reader admits the proposition that for any map between sets, any subset is contained in the preimage of the image of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any adjunction topological space and its any subspace, any subset of the subspace is open if and only if the projections of the preimage of the subset under the quotient map onto the attaching-origin space and the attaching-destination space are open on the projections of the preimage of the subspace under the quotient map with the condition that the attaching-origin space projection accord with the attaching-destination space projection with respect to the attaching map.
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, \(f: S \rightarrow T_2\), the adjunction topological space, \(T_2 \cup_f T_1\) where \(f_2: T_1 \rightarrow T_1 + T_2\) and \(f_3: T_2 \rightarrow T_1 + T_2\) are the inclusions and \(f_4: T_1 + T_2 \rightarrow T_2 \cup_f T_1\) is the quotient map, and any subspace, \(T_3 \subseteq T_2 \cup_f T_1\), any subset, \(S' \subseteq T_3\), is open if and only if \({f_4}^{-1} (S') \cap T_1\) is open on the subspace of \(T_1\), \({T_1}' = {f_4}^{-1} (T_3) \cap T_1\), with an open set, \(U' \subseteq T_1\), such that \({f_4}^{-1} (S') \cap T_1 = U' \cap {T_1}'\) and \({f_4}^{-1} (S') \cap T_2\) is open on the subspace of \(T_2\), \({T_2}' = {f_4}^{-1} (T_3) \cap T_2\), with an open set, \(U'' \subseteq T_2\), such that \({f_4}^{-1} (S') \cap T_2 = U'' \cap {T_2}'\), with the condition that \(U' \cap S = f^{-1} (U'' \cap f (S))\).
2: Proof
Suppose that \({f_4}^{-1} (S') \cap T_1\) is open on \({T_1}'\) and \({f_4}^{-1} (S') \cap T_2\) is open on \({T_2}'\) with the prescribed condition. Let us think of \(U''' := f_4 (f_2 (U') \cup f_3 (U''))\). The condition, \(U' \cap S = f^{-1} (U'' \cap f (S))\), means that for any \([p] \in U'''\) such that \(p \in f (S), [p] = f^{-1} (p) \cup \{p\}\), \(f^{-1} (p) \cup \{p\} \subseteq f_2 (U' \cap S) \cup f_3 (U'' \cap f (S))\), because there is a \(p' \in f_2 (U' \cap S) \cup f_3 (U'' \cap f (S))\) such that \(f_4 (p') = [p]\), which means that \(p' \in f_2 (U' \cap S)\) and \(f (p') = p\) or that \(p' \in f_3 (U'' \cap f (S))\) and \(p' = p\); for the former case, \(p' \in U' \cap S = f^{-1} (U'' \cap f (S))\), \(f (p') = p \in f (f^{-1} (U'' \cap f (S))) \subseteq U'' \cap f (S)\) by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set; for the latter case, \(p = p' \in U'' \cap f (S)\); for any \(p'' \in f^{-1} (p) \cup \{p\}\), \(p'' \in f^{-1} (p)\) or \(p'' = p\); for the former case, \(p'' \in U' \cap S = f_2 (U' \cap S)\) by the condition; for the latter case, \(p'' \in U'' \cap f (S) = f_3 (U'' \cap f (S))\).
The openness of \(U'''\) on \(T_2 \cup_f T_1\) is nothing but the openness of \({f_4}^{-1} (U''') \cap T_1\) on \(T_1\) and the openness of \({f_4}^{-1} (U''') \cap T_2\) on \(T_2\). \({f_4}^{-1} (U''') = f_2 (U') \cup f_3 (U'')\), because while \(f_2 (U') \cup f_3 (U'') \subseteq {f_4}^{-1} (U''')\) is true by the proposition that for any map between sets, any subset is contained in the preimage of the image of the subset, also \({f_4}^{-1} (U''') \subseteq f_2 (U') \cup f_3 (U'')\) holds, because for any \(p\ \in {f_4}^{-1} (U''')\), if \(p \in T_1 \setminus S\) or \(p \in T_2 \setminus f (S)\), \(f_4 (p) = [p] = \{p\} \in U'''\), and there is a \(p' \in f_2 (U') \cup f_3 (U'')\) such that \(f_4 (p') = [p]\), which only \(p' = p\) satisfies, so, \(p \in f_2 (U') \cup f_3 (U'')\); if \(p \in S\) or \(p \in f (S)\), \(f_4 (p) = [p'] \in U'''\) where \(p' \in f (S), [p'] = f^{-1} (p') \cup \{p'\}\), so, by the claim in the last paragraph, \(p \in f^{-1} (p') \cup \{p'\} \subseteq f_2 (U' \cap S) \cup f_3 (U'' \cap f (S)))\). So, \({f_4}^{-1} (U''') \cap T_1 = f_2 (U')\), open on \(T_1\), and \({f_4}^{-1} (U''') \cap T_2 = f_3 (U'')\), open on \(T_2\). So, \(U'''\) is open on \(T_2 \cup_f T_1\).
\(S' = U''' \cap T_3\), because for any \([p] \in S'\), if \(p \in T_1\), \(p \in {f_4}^{-1} (S') \cap T_1 = U' \cap {T_1}' \subseteq U'\), so, \(f_4 (f_2 (p)) = [p] \in U''' = f_4 (f_2 (U') \cup f_3 (U''))\); if \(p \in T_2\), \(p \in {f_4}^{-1} (S') \cap T_2 = U'' \cap {T_2}' \subseteq U''\), so, \(f_4 (f_3 (p)) = [p] \in U''' = f_4 (f_2 (U') \cup f_3 (U''))\); of course \([p] \in T_3\); for any \([p] \in U''' \cap T_3\), if \(p \in T_1\), \(p \in ({f_4}^{-1} (U''') \cap T_1) \cap ({f_4}^{-1} (T_3) \cap T_1) = f_2 (U') \cap {T_1}' = U' \cap {T_1}' = {f_4}^{-1} (S') \cap T_1\), so, \([p] = f_4 (p) \in S'\); if \(p \in T_2\), \(p \in ({f_4}^{-1} (U''') \cap T_2) \cap ({f_4}^{-1} (T_3) \cap T_2) = f_3 (U'') \cap {T_2}' = U'' \cap {T_2}' = {f_4}^{-1} (S') \cap T_2\), so, \([p] = f_4 (p) \in S'\).
So, by the definition of subspace topology, \(S'\) is open.
Suppose that \(S'\) is open. By the definition of subspace topology, there is an open set, \(U''' \subseteq T_2 \cup_f T_1\), such that \(S' = U''' \cap T_3\).
Define \(U': = {f_4}^{-1} (U''') \cap T_1\) and \(U'': = {f_4}^{-1} (U''') \cap T_2\), both open by the definition of quotient topology and the definition of topological sum. They satisfy the condition, \(U' \cap S = f^{-1} (U'' \cap f (S))\), because for any \(p \in U' \cap S\), \(f_4 (p) = [f (p)] \in U'''\), \({f_4}^{-1} ([f (p)]) \cap T_2 = \{f (p)\} \subseteq U''\), so, \(f (p) \in U''\), so, \(p \in f^{-1} (U'') = f^{-1} (U'' \cap f (S))\); for any \(p \in f^{-1} (U'' \cap f (S))\), \(f (p) \in U'' \cap f (S)\), \(f_4 (f (p)) = [p] \in U'''\), so, \(p \in {f_4}^{-1} ([p]) \subseteq {f_4}^{-1} (U''') \cap T_1 = U'\), but as \(p \in S\), \(p \in U' \cap S\).
\({f_4}^{-1} (S') \cap T_1 = U' \cap {T_1}'\), because for any \(p \in {f_4}^{-1} (S') \cap T_1\), \(f_4 (p) = [p] \in S' \subseteq U'''\), so, \(p \in {f_4}^{-1} (U''')\), but as \(p \in T_1\), \(p \in {f_4}^{-1} (U''') \cap T_1 = U'\), and of course, \(p \in {T_1}'\); for any \(p \in U' \cap {T_1}'\), \(f_4 (p) \in f_4 (U' \cap {T_1}') \subseteq f_4 (U') \cap f_4 ({T_1}') \subseteq U''' \cap T_3 = S'\), \(p \in {f_4}^{-1} (S') \cap T_1\). \({f_4}^{-1} (S') \cap T_2 = U'' \cap {T_2}'\), because for any \(p \in {f_4}^{-1} (S') \cap T_2\), \(f_4 (p) = [p] \in S' \subseteq U'''\); \(p \in {f_4}^{-1} (U''') \cap T_2 = U''\); of course, \(p \in {T_2}'\); for any \(p \in U'' \cap {T_2}'\), \(f_4 (p) \in f_4 (U'' \cap {T_2}') \subseteq f_4 (U'') \cap f_4 ({T_2}') \subseteq U''' \cap T_3 = S'\), \(p \in {f_4}^{-1} (S') \cap T_2\).
\(U''' = f_4 (f_2 (U') \cup f_3 (U''))\), because for any \([p] \in U'''\), \(p \in {f_4}^{-1} ([p]) \cap T_1 \subseteq U'\) or \(p \in {f_4}^{-1} ([p]) \cap T_2 \subseteq U''\), so, \([p] \in f_4 (f_2 (U') \cup f_3 (U''))\); for any \([p] \in f_4 (f_2 (U') \cup f_3 (U''))\), \([p] \in f_4 (f_2 (U'))\) or \([p] \in f_4 (f_3 (U''))\), so, there is a \(p \in U'\) such that \(f_4 (f_2 (p)) = [p]\) or a \(p \in U''\) such that \(f_4 (f_3 (p)) = [p]\), so, \([p] \in U'''\).
3: Note
The point is that only the openness of \({f_4}^{-1} (S') \cap T_1\) on \({T_1}'\) and the openness of \({f_4}^{-1} (S') \cap T_2\) on \({T_2}'\) do not guarantee the openness of \(S'\), because \(U'''\) on \(T_2 \cup_f T_1\) is being required.
