description/proof of that topological space is contractible iff space is homotopy equivalent to \(1\)-point topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of homotopy equivalence relation on collection of topological spaces.
- The reader knows a definition of contractible topological space.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is contractible if and only if the space is homotopy equivalent to a \(1\)-point topological space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\sim\): \(= \text{ the homotopy equivalence relation on the collection of the topological spaces }\)
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Statements:
\(T \in \{\text{ the contractible topological spaces }\}\)
\(\iff\)
\(\exists \widetilde{T} \in \{\text{ the } 1 \text{ -point topological spaces }\} (T \sim \widetilde{T})\)
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2: Note
When there is such a \(\widetilde{T}\), \(T\) is homotopy equivalent to each \(1\)-point topological space, because all the \(1\)-point topological spaces are homeomorphic: each \(1\)-point topological space has the inevitable topology, and for each \(1\)-point topological space, \(\widetilde{T}'\), there is the inevitable homeomorphism, \(g: \widetilde{T} \to \widetilde{T}'\), and as \(T \sim \widetilde{T}\), there is a homotopy equivalence, \(f: T \to \widetilde{T}\), which means that \([f]\) is an \(hTop\) isomorphism, which means that there is a continuous \(\widetilde{f}: \widetilde{T} \to T\) such that \([\widetilde{f}] \circ [f] = [id]\) and \([f] \circ [\widetilde{f}] = [id]\), so, there is \(g \circ f: T \to \widetilde{T}'\), and \([\widetilde{f} \circ g^{-1}] \circ [g \circ f] = [\widetilde{f} \circ g^{-1} \circ g \circ f] = [\widetilde{f} \circ f] = [id]\) and \([g \circ f] \circ [\widetilde{f} \circ g^{-1}] = [g \circ f \circ \widetilde{f} \circ g^{-1}] = [g] \circ [f \circ \widetilde{f}] \circ [g^{-1}] = [g] \circ [id] \circ [g^{-1}] = [g \circ id \circ g^{-1}] = [id]\), so, \(T \sim \widetilde{T}'\).
3: Proof
Whole Strategy: Step 1: suppose that \(T\) is contractible; Step 2: see that there is a \(\widetilde{T}\); Step 3: suppose that there is a \(\widetilde{T}\); Step 4: see that \(T\) is contractible.
Step 1:
Let us suppose that \(T\) is contractible.
Step 2:
There is a constant \(f: T \to T, t \mapsto c\) such that \(id \simeq f\), by the definition of contractible topological space.
Let \(\widetilde{T} := \{c\}\), the \(1\)-point topological space with the inevitable topology.
Let \(f': T \to \widetilde{T}, t \mapsto c\).
Let \(g: \widetilde{T} \to T, c \to c\).
\(g \circ f' = f \simeq id\), which means that \([g \circ f'] = [g] \circ [f'] = [id]\).
\(f' \circ g = id \simeq id\), which means that \([f' \circ g] = [f'] \circ [g] = [id]\).
So, \([f']\) is an \(hTop\) isomorphism.
So, \(f'\) is a homotopy equivalence.
So, \(T \sim \widetilde{T}\).
Step 3:
Let us suppose that there is a \(\widetilde{T}\) such that \(T \sim \widetilde{T}\).
Step 4:
Let \(\widetilde{T} = \{c\}\).
There is a homotopy equivalence, \(h: T \to \widetilde{T}\).
So, there is a continuous map, \(g: \widetilde{T} \to T\), such that \([g] \circ [h] = [id]\) and \([h] \circ [g] = [id]\).
\(f := g \circ h: T \to T\) is constant, because for each \(t \in T\), \(g \circ h (t) = g (c)\).
\([f] = [g \circ h] = [g] \circ [h] = [id]\), which means that \(f \simeq id\).
So, \(T\) is contractible.
