description/proof of that adjunction topological space is connected if domain of attaching map is nonempty and attaching-origin space and attaching-destination space are connected
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 connected topological space.
- The reader admits the proposition that for any topological sum, each constituent is the topological subspace of the topological sum.
- 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.
- The reader admits the proposition that the preimage of the whole codomain of any map is the whole domain.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
Target Context
- The reader will have a description and a proof of the proposition that any adjunction topological space is connected if the domain of the attaching map is nonempty and the attaching-origin space and the attaching-destination 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the connected topological spaces }\}\)
\(T_2\): \(\in \{\text{ the connected topological spaces }\}\)
\(S\): \(\subseteq T_1\), such that \(S \neq \emptyset\)
\(f\): \(: S \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(T_1 + T_2\): \(= \text{ the topological sum }\)
\(T_2 \cup_f T_1\): \(= \text{ the adjunction topological space }\)
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Statements:
\(T_2 \cup_f T_1 \in \{\text{ the connected topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(g \vert_{T_1}: T_1 \to T_2 \cup_f T_1\) and \(g \vert_{T_2}: T_2 \to T_2 \cup_f T_1\) are continuous; Step 2: suppose that \(T_2 \cup_f T_1\) was not connected; Step 3: find a contradiction.
Step 1:
Let \(g: T_1 + T_2 \to T_2 \cup_f T_1\) be the one defined in the definition of adjunction topological space obtained by attaching topological space via map to topological space.
\(g\) is continuous: the topology of \(T_2 \cup_f T_1\) is defined to make \(g\) continuous.
\(T_1\) is the topological subspace of \(T_1 + T_2\), by the proposition that for any topological sum, each constituent is the topological subspace of the topological sum; \(T_2\) is the topological subspace of \(T_1 + T_2\), likewise.
\(g \vert_{T_1}: T_1 \to T_2 \cup_f T_1\) and \(g \vert_{T_2}: T_2 \to T_2 \cup_f T_1\) are continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 2:
Let us suppose that \(T_2 \cup_f T_1\) was not connected.
Step 3:
\(T_2 \cup_f T_1 = U_1 \cup U_2\) where \(U_j \subseteq T_2 \cup_f T_1\) would be nonempty open and \(U_1 \cap U_2 = \emptyset\).
\({g \vert_{T_1}}^{-1} (U_1 \cup U_2) = {g \vert_{T_1}}^{-1} (U_1) \cup {g \vert_{T_1}}^{-1} (U_2) = T_1\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets and the proposition that the preimage of the whole codomain of any map is the whole domain.
\({g \vert_{T_1}}^{-1} (U_1) \cap {g \vert_{T_1}}^{-1} (U_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
As \(g \vert_{T_1}\) was continuous, \({g \vert_{T_1}}^{-1} (U_j)\) would be open on \(T_1\).
If \({g \vert_{T_1}}^{-1} (U_1) \neq \emptyset\) and \({g \vert_{T_1}}^{-1} (U_2) \neq \emptyset\), it would be a contradiction against \(T_1\)'s being connected.
Let us suppose that \({g \vert_{T_1}}^{-1} (U_2) = \emptyset\) without loss of generality.
Then, \({g \vert_{T_2}}^{-1} (U_2) \neq \emptyset\), because otherwise, \(g^{-1} (U_2) = \emptyset\), a contradiction against that \(g\) was a surjection and \(U_2 \neq \emptyset\).
\({g \vert_{T_2}}^{-1} (U_1 \cup U_2) = {g \vert_{T_2}}^{-1} (U_1) \cup {g \vert_{T_2}}^{-1} (U_2) = T_2\), as before.
\({g \vert_{T_2}}^{-1} (U_1) \cap {g \vert_{T_2}}^{-1} (U_2) = \emptyset\), as before.
\({g \vert_{T_2}}^{-1} (U_j)\) would be open on \(T_2\), as before.
\({g \vert_{T_2}}^{-1} (U_1) \neq \emptyset\), because otherwise, \(U_1 \subseteq g \vert_{T_1} (T_1 \setminus S)\), because for each \(u_1 \in U_1\), while \(u_1 \in g \vert_{T_1} (T_1)\), which would mean that \(u_1 = g \vert_{T_1} (t_1)\) for an \(t_1 \in T_1\), \(t_1 \notin S\), because otherwise, \(u_1 = g \vert_{T_1} (t_1) = g (f (t_1)) = g \vert_{T_2} (f (t_1))\), a contradiction, and \(U_2 \subseteq g \vert_{T_2} (T_2 \setminus f (S))\), because for each \(u_2 \in U_2\), while \(u_2 \in g \vert_{T_2} (T_2)\), which would mean that \(u_2 = g \vert_{T_2} (t_2)\) for an \(t_2 \in T_2\), \(t_2 \notin f (S)\), because otherwise, \(t_2 = f (s)\) for an \(s \in S\) and \(u_2 = g \vert_{T_2} (t_2) = g (f (s)) = g \vert_{T_1} (s)\), a contradiction, then, \(g (S)\), nonempty as \(S\) is nonempty, would be disjoint from \(U_1 \cup U_2\), because for each \(g (s) \in g (S)\), \(g (s) \notin g \vert_{T_1} (T_1 \setminus S)\), because if \(g (s) = g (t_1)\) for an \(t_1 \in T_1\), \(s = t_1\) or \(f (s) = f (t_1)\) was necessary, which would mean that \(t_1 \in S\), and \(g (s) \notin g \vert_{T_2} (T_2 \setminus f (S))\), because if \(g (s) = g (t_2)\) for an \(t_2 \in T_2\), \(f (s) = t_2\) was necessary, which would mean that \(t_2 \in f (S)\), a contradiction.
That would be a contradiction against \(T_2\)'s being connected.
So, \(T_2 \cup_f T_1\) is connected.
