description/proof of that contractible topological space is path-connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of contractible topological space.
- The reader knows a definition of path-connected topological space.
- The reader admits the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed 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 contractible topological space is 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:
\(T\): \(\in \{\text{ the contractible topological spaces }\}\)
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Statements:
\(T \in \{\text{ the path-connected topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take a constant map, \(f: T \to T, t \mapsto c\), such that \(f \simeq id\) and take a homotopy, \(f': T \times I \to T\), and see that \(f' (t', ): I \to T\) is a path from \(t'\) to \(c\); Step 2: see that for each \(t', t'' \in T\), there is a path from \(t'\) to \(t''\).
Step 1:
There is a constant map, \(f: T \to T, t \mapsto c\), such that \(f \simeq id\), by the definition of contractible topological space.
So, there is a homotopy, \(f': T \times I \to T\), such that for each \(t \in T\), \(f' (t, 0) = id (t) = t\) and \(f' (t, 1) = f (t) = c\).
Let \(t' \in T\) be any.
Let \(f' (t', ): I \to T\) be the map induced from \(f'\) with \(t'\) fixed.
\(f' (t', )\) is continuous, by the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
\(f' (t', 0) = t'\) and \(f' (t', 1) = c\).
So, \(f' (t', )\) is a path from \(t'\) to \(c\).
Step 2:
Let \(t', t'' \in T\) be any.
By Step 1, there are a path from \(t'\) to \(c\), \(\lambda': I \to T\), and a path from \(t''\) to \(c\), \(\lambda'': I \to T\).
Let us take \(\widetilde{\lambda''}: [1, 2] \to T, j \mapsto \lambda'' (1 - (j - 1))\), which is valid, because \(1 - (j - 1) \in [0, 1]\), because \(j - 1 \in [0, 1]\), \(- (j - 1) \in [- 1, 0]\), and \(1 - (j - 1) \in [0, 1]\).
\(\widetilde{\lambda''} = \lambda'' \circ g\) where \(g: [1, 2] \to [0, 1], j \to 1 - (j - 1)\).
\(g\) is obviously continuous, so, \(\widetilde{\lambda''}\) 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.
Let us take \(\lambda: [0, 2] \to T, j \mapsto \lambda' (j) \text{ when } j \in [0, 1]; \mapsto \widetilde{\lambda''} (j) \text{ when } j \in (1, 2]\).
\(\lambda \vert_{[0, 1]} = \lambda'\).
\(\lambda \vert_{[1, 2]} = \widetilde{\lambda''}\), because \(\lambda \vert_{[1, 2]} (1) = \lambda' (1) = c = \lambda'' (1) = \lambda'' (1 - (1 - 1)) = \widetilde{\lambda''} (1)\).
\(\lambda\) 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.
\(\lambda (0) = t'\) and \(\lambda (2) = t''\).
So, \(\lambda\) is a path from \(t'\) to \(t''\).
So, \(T\) is path-connected.
