description/proof of that adjunction topological space is path-connected if domain of attaching map is nonempty and attaching-origin space and attaching-destination space are path-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 path-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 maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any adjunction topological space is path-connected if the domain of the attaching map is nonempty and the attaching-origin space and the attaching-destination space are path-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:
\(T_1\): \(\in \{\text{ the path-connected topological spaces }\}\)
\(T_2\): \(\in \{\text{ the path-connected topological spaces }\}\)
\(S\): \(\subseteq T_1\), 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:
\(T_2 \cup_f T_1 \in \{\text{ the path-connected topological spaces }\}\)
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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: see that for each \(t, t' \in T_2 \cup_f T_1\), there are \(3\) nonexclusive cases; Step 3: deal with the case 1); Step 4: deal with the case 2); Step 5: deal with the case 3); Step 6: conclude the proposition.
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:
For each \(t, t' \in T_2 \cup_f T_1\), there are \(3\) nonexclusive cases: 1) \(t, t' \in g \vert_{T_1} (T_1)\); 2) \(t, t' \in g \vert_{T_2} (T_2)\); 3) \(t \in g \vert_{T_1} (T_1)\) and \(t' \in g \vert_{T_2} (T_2)\) without loss of generality.
They are nonexclusive because \(t\) or \(t'\) may be both in \(g \vert_{T_1} (T_1)\) and \(g \vert_{T_2} (T_2)\), but that does not matter, because the point is that any case can be regarded to be of at least one of the \(3\) cases.
Step 3:
Let us deal with the case 1).
There are some points, \(t_1, t'_1 \in T_1\), such that \(t = g \vert_{T_1} (t_1)\) and \(t' = g \vert_{T_1} (t'_1)\).
There is a path, \(\gamma_1: [0, 1] \to T_1\), such that \(\gamma_1 (0) = t_1\) and \(\gamma_1 (1) = t'_1\), because \(T_1\) is path-connected.
There is the path, \(g \vert_{T_1} \circ \gamma_1: [0, 1] \to T_2 \cup_f T_1\), which is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point, and satisfies \(g \vert_{T_1} \circ \gamma_1 (0) = t\) and \(g \vert_{T_1} \circ \gamma_1 (1) = t'\).
Step 4:
Let us deal with the case 2).
There are some points, \(t_2, t'_2 \in T_2\), such that \(t = g \vert_{T_2} (t_2)\) and \(t' = g \vert_{T_2} (t'_2)\).
There is a path, \(\gamma_2: [0, 1] \to T_2\), such that \(\gamma_2 (0) = t_2\) and \(\gamma_2 (1) = t'_2\), because \(T_2\) is path-connected.
There is the path, \(g \vert_{T_2} \circ \gamma_2: [0, 1] \to T_2 \cup_f T_1\), which is continuous, as before, and satisfies \(g \vert_{T_2} \circ \gamma_2 (0) = t\) and \(g \vert_{T_2} \circ \gamma_2 (1) = t'\).
Step 5:
Let us deal with the case 3).
There are some points, \(t_1 \in T_1\) and \(t_2 \in T_2\), such that \(t = g \vert_{T_1} (t_1)\) and \(t' = g \vert_{T_2} (t_2)\).
As \(S \neq \emptyset\), there is an \(s \in S\).
There is a path, \(\gamma_1: [0, 0.5] \to T_1\), such that \(\gamma_1 (0) = t_1\) and \(\gamma_1 (0.5) = s\), because \(T_1\) is path-connected.
There is a path, \(\gamma_2: [0.5, 1] \to T_2\), such that \(\gamma_2 (0.5) = f (s)\) and \(\gamma_2 (1) = t_2\), because \(T_2\) is path-connected.
There is the path, \(g \vert_{T_1} \circ \gamma_1: [0, 0.5] \to T_2 \cup_f T_1\), which is continuous, as before, and satisfies \(g \vert_{T_1} \circ \gamma_1 (0) = t\) and \(g \vert_{T_1} \circ \gamma_1 (0.5) = g \vert_{T_1} (s)\).
There is the path, \(g \vert_{T_2} \circ \gamma_2: [0.5, 1] \to T_2 \cup_f T_1\), which is continuous, as before, and satisfies \(g \vert_{T_2} \circ \gamma_2 (0.5) = g \vert_{T_2} (f (s))\) and \(g \vert_{T_2} \circ \gamma_2 (1) = t'\).
But \(g \vert_{T_1} (s) = g \vert_{T_2} (f (s))\).
Then, \(\gamma: [0, 1] \to T_2 \cup_f T_1\) where \(\gamma\vert_{[0, 0.5]} = \gamma_1\) and \(\gamma\vert_{[0.5, 1]} = \gamma_2\) is well-defined as \(\gamma_1 (0.5) = \gamma_2 (0.5)\) and is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous, and satisfies \(\gamma (0) = t\) and \(\gamma (1) = t'\).
Step 6:
So, for each case, there is a path that connects \(t\) and \(t'\).
So, \(T_2 \cup_f T_1\) is path-connected.
