2024-12-15

899: On Set of Continuous Maps Between Topological Spaces, Being Homotopic Is Equivalence Relation

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description/proof of that on set of continuous maps between topological spaces, being homotopic is equivalence relation

Topics


About: topological space

The table of contents of this article


Starting Context



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, being homotopic 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(= \{f: T_1 \to T_2: f \in \{\text{ the continuous maps }\}\}\)
\(\sim\): \(\subseteq S \times S\), \(\in \{\text{ the relations }\}\), such that \(\forall f_1, f_2 \in S (f_1 \sim f_2 \iff f_1 \simeq f_2)\), where \(\simeq\) means being homotopic
//

Statements:
\(\sim \in \{\text{ the equivalence relations }\}\)
//


2: 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 \(F (t, 1) = f (t)\).

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 \(F (t, 1) = f_2 (t)\); let \(F': T_1 \times I \to T_2, (t, r) \mapsto F (t, 1 - r)\), which is continuous, because \(F' = F \circ (id, g)\) where \(g: I \to I, r \mapsto 1 - r\) 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 \(F' (t, 1) = F (t, 0) = f_1 (t)\).

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 \(F_1 (t, 0) = f_1 (t)\) and \(F_1 (t, 1) = f_2 (t)\) and a continuous \(F_2: T_1 \times I \to T_2\) such that \(F_2 (t, 0) = f_2 (t)\) and \(F_2 (t, 1) = f_3 (t)\); let us define \(F_3: T_1 \times I \to T_2\) as on \(T_1 \times [0, 1 / 2]\), \(= F_1 (t, 2 r)\), and on \(T_1 \times [1 / 2, 1]\), \(= F_2 (t, 2 (r - 1 / 2))\), 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 \(F_1 (t, 2 r) = F_1 \circ (id, g)\) where \(g: [0, 1 / 2] \to [0, 1], r \mapsto 2 r\) and \(F_2 (t, 2 (r - 1 / 2)) = F_2 \circ (id, h)\) where \(h: [1 / 2, 1] \to [0, 1], r \to 2 (r - 1 / 2)\), 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; \(F_3 (t, 0) = F_1 (t, 0) = f_1 (t)\) and \(F_3 (t, 1) = F_2 (t, 1) = f_3 (t)\).


References


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