description/proof of that for homotopy equivalence, map homotopic to homotopy equivalence is homotopy equivalence
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of homotopy equivalence.
- The reader knows a definition of homotopic maps.
- 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 from any topological space into any product topological space is continuous if and only if each component map is continuous.
- The reader admits the proposition that for any homotopic maps from any 1st topological space into any 2nd topological space and any homotopic maps from the 2nd topological space into any 3rd topological space, the compositions of the homotopic maps are homotopic with a homotopy as this.
- The reader admits the proposition that on the set of the continuous maps between any topological spaces, being homotopic is an equivalence relation.
Target Context
- The reader will have a description and a proof of the proposition that for any homotopy equivalence, any map homotopic to the homotopy equivalence is a homotopy equivalence.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the homotopy equivalences from } T_1 \text{ into } T_2\}\)
\(f'_1\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(f_1 \simeq f'_1\)
\(\implies\)
\(f'_1 \in \{\text{ the homotopy equivalences from } T_1 \text{ into } T_2\}\)
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2: Proof
Whole Strategy: Step 1: take any continuous map, \(f_2: T_2 \to T_1\), such that \(f_2 \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ f_2 \simeq id_{T_2}\) and a homotopy from \(f_1\) to \(f'_1\), \(F_1\); Step 2: take any homotopy from \(f_2 \circ f_1\) to \(id_{T_1}\), \(F\), and \(F': T_1 \times I \to T_1, (t_1, j) \mapsto F (f_2 \circ F_1 (t_1, j), j)\), and see that \(F'\) is a homotopy from \(f_2 \circ f_1 \circ f_2 \circ f_1\) to \(f_2 \circ f'_1\) and that \(id_{T_1} \simeq f_2 \circ f'_1\); Step 3: take any homotopy from \(f_1 \circ f_2\) to \(id_{T_2}\), \(F\), and \(F': T_2 \times I \to T_2, (t_2, j) \mapsto F (F_1 (f_2 (t_2), j), j)\), and see that \(F'\) is a homotopy from \(f_1 \circ f_2 \circ f_1 \circ f_2\) to \(f'_1 \circ f_2\) and that \(id_{T_2} \simeq f'_1 \circ f_2\); Step 4: conclude the proposition.
Step 1:
There is a continuous map, \(f_2: T_2 \to T_1\), such that \(f_2 \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ f_2 \simeq id_{T_2}\), by Note for the definition of homotopy equivalence.
There is a homotopy from \(f_1\) to \(f'_1\), \(F_1: T_1 \times I \to T_2\): for each \(t_1 \in T_1\), \(F_1 (t_1, 0) = f_1 (t_1)\) and \(F_1 (t_1, 1) = f'_1 (t_1)\).
Step 2:
Let \(F: T_1 \times I \to T_1\) be any homotopy from \(f_2 \circ f_1\) to \(id_{T_1}\): for each \(t_1 \in T_1\), \(F (t_1, 0) = f_2 \circ f_1 (t_1)\) and \(F (t_1, 1) = id_{T_1} (t_1)\).
Let us take \(F': T_1 \times I \to T_1, (t_1, j) \mapsto F (f_2 \circ F_1 (t_1, j), j)\).
\(F'': T_1 \times I \to T_1 \times I, (t_1, j) \mapsto (f_2 \circ F_1 (t_1, j), j)\) 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 the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous: \(: T_1 \times I \to I, (t_1, j) \mapsto j\) is continuous, because for each open neighborhood of \(j\), \(U_j \subseteq I\), \(T_1 \times U_j\) is mapped into \(U_j\) where \(T_1 \times U_j \subseteq T_1 \times I\) is an open neighborhood of \((t, j)\).
\(F' = F \circ F''\) 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.
For each \(t_1 \in T_1\), \(F' (t_1, 0) = F (f_2 \circ F_1 (t_1, 0), 0) = F (f_2 \circ f_1 (t_1), 0) = f_2 \circ f_1 (f_2 \circ f_1 (t_1)) = f_2 \circ f_1 \circ f_2 \circ f_1 (t_1)\) and \(F' (t_1, 1) = F (f_2 \circ F_1 (t_1, 1), 1) = F (f_2 \circ f'_1 (t_1), 1) = id_{T_1} (f_2 \circ f'_1 (t_1)) = f_2 \circ f'_1 (t_1)\).
So, \(f_2 \circ f_1 \circ f_2 \circ f_1 \simeq f_2 \circ f'_1\).
But as \(id_{T_1} \simeq f_2 \circ f_1\), \(id_{T_1} \circ f_2 \circ f_1 \simeq f_2 \circ f_1 \circ f_2 \circ f_1\), by the proposition that for any homotopic maps from any 1st topological space into any 2nd topological space and any homotopic maps from the 2nd topological space into any 3rd topological space, the compositions of the homotopic maps are homotopic with a homotopy as this, but the left hand side is \(f_2 \circ f_1\), so, \(f_2 \circ f_1 \simeq f_2 \circ f_1 \circ f_2 \circ f_1\).
So, \(id_{T_1} \simeq f_2 \circ f_1 \simeq f_2 \circ f_1 \circ f_2 \circ f_1 \simeq f_2 \circ f'_1\).
By the proposition that on the set of the continuous maps between any topological spaces, being homotopic is an equivalence relation, \(id_{T_1} \simeq f_2 \circ f'_1\).
Step 3:
Let \(F: T_2 \times I \to T_2\) be any homotopy from \(f_1 \circ f_2\) to \(id_{T_2}\): for each \(t_2 \in T_2\), \(F (t_2, 0) = f_1 \circ f_2 (t_2)\) and \(F (t_2, 1) = id_{T_2} (t_2)\).
Let us take \(F': T_2 \times I \to T_2, (t_2, j) \mapsto F (F_1 (f_2 (t_2), j), j)\).
\(F'': T_2 \times I \to T_2 \times I, (t_2, j) \mapsto (F_1 (f_2 (t_2), j), j)\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies, 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 the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous: \(: T_1 \times I \to I, (t_1, j) \mapsto j\) is continuous, because for each open neighborhood of \(j\), \(U_j \subseteq I\), \(T_1 \times U_j\) is mapped into \(U_j\) where \(T_1 \times U_j \subseteq T_1 \times I\) is an open neighborhood of \((t, j)\).
\(F' = F \circ F''\) 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.
For each \(t_2 \in T_2\), \(F' (t_2, 0) = F (F_1 (f_2 (t_2), 0), 0) = F (f_1 (f_2 (t_2)), 0) = f_1 \circ f_2 (f_1 (f_2 (t_2))) = f_1 \circ f_2 \circ f_1 \circ f_2 (t_2)\) and \(F' (t_2, 1) = F (F_1 (f_2 (t_2), 1), 1) = F (f'_1 (f_2 (t_2)), 1) = id_{T_2} (f'_1 (f_2 (t_2))) = f'_1 (f_2 (t_2)) = f'_1 \circ f_2 (t_2)\).
So, \(f_1 \circ f_2 \circ f_1 \circ f_2 \simeq f'_1 \circ f_2\).
But as \(id_{T_2} \simeq f_1 \circ f_2\), \(id_{T_2} \circ f_1 \circ f_2 \simeq f_1 \circ f_2 \circ f_1 \circ f_2\), by the proposition that for any homotopic maps from any 1st topological space into any 2nd topological space and any homotopic maps from the 2nd topological space into any 3rd topological space, the compositions of the homotopic maps are homotopic with a homotopy as this, but the left hand side is \(f_1 \circ f_2\), so, \(f_1 \circ f_2 \simeq f_1 \circ f_2 \circ f_1 \circ f_2\).
So, \(id_{T_2} \simeq f_1 \circ f_2 \simeq f_1 \circ f_2 \circ f_1 \circ f_2 \simeq f'_1 \circ f_2\).
By the proposition that on the set of the continuous maps between any topological spaces, being homotopic is an equivalence relation, \(id_{T_2} \simeq f'_1 \circ f_2\).
Step 4:
By Step 2 and Step 3, \(f'_1\) is a homotopy equivalence, by Note for the definition of homotopy equivalence.
