description/proof of that on set of continuous maps between topological spaces, for subset of domain, being homotopic relative to subset is equivalence relation
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of maps homotopic relative to subset of domain.
- The reader knows a definition of equivalence relation on set.
- The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
- 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 on the set of the continuous maps between any topological spaces, for any subset of the domain, being homotopic relative to the subset is an equivalence relation.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(= \{f: T_1 \to T_2: f \in \{\text{ the continuous maps }\}\}\)
\(S_1\): \(\subseteq T_1\)
\(\sim\): \(\subseteq S \times S\), \(\in \{\text{ the relations }\}\), such that \(\forall f_1, f_2 \in S (f_1 \sim f_2 \iff f_1 \simeq f_2 rel S_1)\), where \(\simeq rel\) means being homotopic relative
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Statements:
\(\sim \in \{\text{ the equivalence relations }\}\)
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2: Note
For each \(f, f' \in S\) such that \(f \sim f'\), \(f (S_1) = f' (S_1)\), because for each \(s_1 \in S_1\), \(f (s_1) = f' (s_1)\).
For any \(S_2 \subseteq T_2\), \(S^` := \{f: T_1 \to T_2: f \in \{\text{ the continuous maps }\} \vert f (S_1) = S_2\}\) is a subset of \(S\), and \(S^`\) has the subset equivalence relation.
\(S^`\) selects some equivalence classes of \(S\), which means that for each \(f \in S^`\), \([f] \subseteq S^`\).
3: Proof
Whole Strategy: Step 1: see that \(\sim\) satisfies the 3 requirements to be an equivalence relation.
Step 1:
1) \(\forall f \in S (f \sim f)\): reflexivity: let \(F: T_1 \times I \to T_2, (t, r) \mapsto f (t)\), which is continuous, because as \(f\) is continuous, for each open neighborhood of \(f (t)\), \(U_{f (t)} \subseteq T_2\), there is an open neighborhood of \(t\), \(U_t \subseteq T_1\), such that \(f (U_t) \subseteq U_{f (t)}\), and \(U_t \times I \subseteq T_1 \times I\) is an open neighborhood of \((t, r)\) and \(F (U_t \times I) \subseteq U_{f (t)}\); \(F (t, 0) = f (t)\) and \(F (t, 1) = f (t)\); for each \(s_1 \in S_1\) and each \(j \in I\), \(F (s_1, j) = f (s_1) = f (s_1)\).
2) \(\forall f_1, f_2 \in S (f_1 \sim f_2 \implies f_2 \sim f_1)\): symmetry: there is a continuous \(F: T_1 \times I \to T_2\) such that \(F (t, 0) = f_1 (t)\) and \(F (t, 1) = f_2 (t)\) and \(F (s_1, j) = f_1 (s_1) = f_2 (s_1)\) for each \(s_1 \in S_1\) and each \(j \in I\); let \(F': T_1 \times I \to T_2, (t, j) \mapsto F (t, 1 - j)\), which is continuous, because \(F' = F \circ (id, g)\) where \(g: I \to I, j \mapsto 1 - j\) and \(id\) and \(g\) are obviously continuous and \((id, g)\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies, and the composition 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; \(F' (t, 0) = F (t, 1) = f_2 (t)\) and \(F' (t, 1) = F (t, 0) = f_1 (t)\) and \(F' (s_1, j) = F (s_1, 1 - j) = f_1 (s_1) = f_2 (s_1)\) for each \(s_1 \in S_1\) and each \(j \in I\).
3) \(\forall f_1, f_2, f_3 \in S ((f_1 \sim f_2 \land f_2 \sim f_3)\implies f_1 \sim f_3)\): transitivity: there are a continuous \(F_1: T_1 \times I \to T_2\) such that \(F_1 (t, 0) = f_1 (t)\), \(F_1 (t, 1) = f_2 (t)\), and \(\forall s_1 \in S_1 (F_1 (s_1, j) = f_1 (s_1) = f_2 (s_1))\) and a continuous \(F_2: T_1 \times I \to T_2\) such that \(F_2 (t, 0) = f_2 (t)\), \(F_2 (t, 1) = f_3 (t)\), and \(\forall s_1 \in S_1 (F_2 (s_1, j) = f_2 (s_1) = f_3 (s_1))\); let us define \(F_3: T_1 \times I \to T_2\) as over \(T_1 \times [0, 1 / 2]\), \(= F_1 (t, 2 j)\), and over \(T_1 \times [1 / 2, 1]\), \(= F_2 (t, 2 (j - 1 / 2))\), which is well-defined, because while \(\{T_1 \times [0, 1 / 2], T_1 \times [1 / 2, 1]\}\) is a closed cover of \(T_1 \times I\), \(F_3\) is consistent because \(F_3 (t, 1 / 2) = F_1 (t, 1) = f_2 (t) = F_2 (t, 0)\), and \(F_3\) is continuous on \(T_1 \times [0, 1 / 2]\) and \(T_1 \times [1 / 2, 1]\), because \(F_1 (t, 2 j) = F_1 \circ (id, g)\) where \(g: [0, 1 / 2] \to [0, 1], j \mapsto 2 j\) and \(F_2 (t, 2 (j - 1 / 2)) = F_2 \circ (id, h)\) where \(h: [1 / 2, 1] \to [0, 1], j \to 2 (j - 1 / 2)\), and \(F_3\) 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; \(F_3 (t, 0) = F_1 (t, 0) = f_1 (t)\), \(F_3 (t, 1) = F_2 (t, 1) = f_3 (t)\), and for each \(s_1 \in S_1\), for each \(j \in [0, 1 / 2]\), \(F_3 (s_1, j) = F_1 (s_1, 2 j) = f_1 (s_1) = f_2 (s_1) = f_3 (s_1)\) and for \(j \in [1 / 2, 1]\), \(F_3 (s_1, j) = F_2 (s_1, 2 (j - 1 / 2)) = f_2 (s_1) = f_3 (s_1)\), but \(f_1 (s_1) = f_2 (s_1)\), so, for each \(j \in I\), \(F_3 (s_1, j) = f_1 (s_1) = f_3 (s_1)\).
