<The previous article in this series | The table of contents of this series |
description/proof of that intersection of non-increasing sequence of nonempty open or closed subsets does not necessarily contain point
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 the intersection of a non-increasing sequence of nonempty open or closed subsets does not necessarily contain a point.
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 1
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{\text{ the open subsets of } T\} \setminus \{\emptyset\}\), \(\in \{\text{ the sequences }\}\), such that \(\forall j, j' \in J \text{ such that } j \lt j' (s (j') \subseteq s (j))\)
//
Statements:
not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)"
//
2: Proof 1
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let us see a counterexample.
Let \(T := \mathbb{R}\), as the Euclidean topological space, \(J = \mathbb{N}\), and \(s (j) := (0, 1 / (j + 1)) \subseteq T\).
\(s\) satisfies the conditions, because for each \(j \in J\), \(s (j) \neq \emptyset\), and for each \(j, j' \in J\) such that \(j \lt j'\), \(s (j') = (0, 1 / (j' + 1)) \subseteq (0, 1 / (j + 1)) = s (j)\).
Let \(t \in T\) be any.
When \(t \le 0\), \(t \notin (0, 1) = s (0)\), so, \(t \notin \cap_{j \in J} s (j)\).
When \(0 \lt t\), there is a large enough \(j \in J\) such that \(1 / (j + 1) \lt t\), so, \(t \notin (0, 1 / (j + 1)) = s (j)\), so, \(t \notin \cap_{j \in J} s (j)\).
So, anyway, \(t \notin \cap_{j \in J} s (j)\).
So, \(\cap_{j \in J} s (j) = \emptyset\).
3: Structured Description 2
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{\text{ the closed subsets of } T\} \setminus \{\emptyset\}\), \(\in \{\text{ the sequences }\}\), such that \(\forall j, j' \in J \text{ such that } j \lt j' (s (j') \subseteq s (j))\)
//
Statements:
not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)"
//
4: Proof 2
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let us see a counterexample.
Let \(T := \mathbb{R}\), as the Euclidean topological space, \(J = \mathbb{N}\), and \(s (j) := (- \infty, - j] \subseteq T\).
\(s\) satisfies the conditions, because for each \(j \in J\), \(s (j) \neq \emptyset\), and for each \(j, j' \in J\) such that \(j \lt j'\), \(s (j') = (- \infty, - j'] \subseteq (- \infty, - j] = s (j)\).
Let \(t \in T\) be any.
There is a large enough \(j \in J\) such that \(- j \lt t\), so, \(t \notin (- \infty, - j] = s (j)\), so, \(t \notin \cap_{j \in J} s (j)\).
So, \(t \notin \cap_{j \in J} s (j)\).
So, \(\cap_{j \in J} s (j) = \emptyset\).
5: Note
When the codomain of \(s\) is not restricted to the set of the nonempty open subsets or the set of the nonempty closed subsets but to the set of the nonempty subsets, not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)" even more, because Proof 1 and Proof 2 are some counterexamples.
The point of Description 2 is that \(T\) is not necessarily compact.
If \(T\) is compact, the intersection necessarily contains a point, by the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that union of complements of subsets is whole set iff intersection of subsets is empty
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any set, the union of the complements of any possibly uncountable number of subsets is the whole set if and only if the intersection of the subsets is empty.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S \vert j \in J\}\):
//
Statements:
\(\cup_{j \in J} (S \setminus S_j) = S\)
\(\iff\)
\(\cap_{j \in J} S = \emptyset\)
//
2: Proof
Whole Strategy: Step 1: see that \(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\); Step 2: suppose that \(\cup_{j \in J} (S \setminus S_j) = S\); Step 3: see that \(\cap_{j \in J} S = \emptyset\); Step 4: suppose that \(\cap_{j \in J} S = \emptyset\); Step 5: see that \(\cup_{j \in J} (S \setminus S_j) = S\).
Step 1:
\(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\), the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
Step 2:
Let us suppose that \(\cup_{j \in J} (S \setminus S_j) = S\).
Step 3:
\(S \setminus \cap_{j \in J} S = \cup_{j \in J} (S \setminus S_j)\), by Step 1, \(= S\), by the supposition, which implies that \(\cap_{j \in J} S = \emptyset\).
Step 4:
Let us suppose that \(\cap_{j \in J} S = \emptyset\).
Step 5:
\(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\), by Step 1, \(= S \setminus \emptyset\), by the supposition, \(= S\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that composition of homotopy equivalences is homotopy equivalence
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 the composition of any homotopy equivalences 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the homotopy equivalences }\}\)
\(f_2\): \(: T_2 \to T_3\), \(\in \{\text{ the homotopy equivalences }\}\)
//
Statements:
\(f_2 \circ f_1 \in \{\text{ the homotopy equivalences }\}\)
//
2: Note
Some people may think that this proposition is already included in the fact that being homotopy equivalent is an equivalence relation, but that fact implies only that there is a homotopy equivalence from \(T_1\) into \(T_3\), not \(f_2 \circ f_1\) is a homotopy equivalence: certainly, being an equivalence relation is usually proved by proving that \(f_2 \circ f_1\) is a homotopy equivalence, though.
3: Proof
Whole Strategy: Step 1: take a continuous \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\) and a continuous \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\); Step 2: see that \(\widetilde{f_2} \circ \widetilde{f_3} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{f_2} \circ \widetilde{f_3} \simeq id_{T_3}\).
Step 1:
There is a continuous \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\), by Note for the definition of homotopy equivalence.
There is a continuous \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\), as before.
Step 2:
Let us take \(\widetilde{f_2} \circ \widetilde{f_3}: T_3 \to T_1\).
\(\widetilde{f_2} \circ \widetilde{f_3}\) 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.
\(\widetilde{f_2} \circ \widetilde{f_3} \circ f_2 \circ f_1 = \widetilde{f_2} \circ (\widetilde{f_3} \circ f_2) \circ f_1 \simeq \widetilde{f_2} \circ id_{T_2} \circ f_1\), because \(\widetilde{f_3} \circ f_2 \simeq T_2\), 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, \(= \widetilde{f_2} \circ f_1 \simeq id_{T_1}\).
\(f_2 \circ f_1 \circ \widetilde{f_2} \circ \widetilde{f_3} = f_2 \circ (f_1 \circ \widetilde{f_2}) \circ \widetilde{f_3} \simeq f_2 \circ id_{T_2} \circ \widetilde{f_3}\), because \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\), as before, \(= f_2 \circ \widetilde{f_3} \simeq id_{T_3}\).
So, \(f_2 \circ f_1\) is a homotopy equivalence.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for continuous map from 1st space into 2nd space and continuous map from 2nd space into 3rd space, if 1st map and composition of 2nd map after 1st map are homotopy equivalences, 2nd map is homotopy equivalence
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 for any continuous map from any 1st topological space into any 2nd topological space and any continuous map from the 2nd space into any 3rd space, if the 1st map and the composition of the 2nd map after the 1st map are some homotopy equivalences, the 2nd map 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(f_2\): \(: T_2 \to T_3\), \(\in \{\text{ the continuous maps }\}\)
//
Statements:
\(f_1 \in \{\text{ the homotopy equivalences }\} \land f_2 \circ f_1 \in \{\text{ the homotopy equivalences }\}\)
\(\implies\)
\(f_2 \in \{\text{ the homotopy equivalences }\}\)
//
2: Note
\(f_2\) needs to be presupposed to be continuous, for this proposition, because Proof does not prove that \(f_2\) is continuous, which is a necessity for \(f_2\) to be a homotopy equivalence.
3: Proof
Whole Strategy: Step 1: take \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\) and \(\widetilde{\widetilde{f_3}}: T_3 \to T_1\) such that \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\); Step 2: see that \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\) and \(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \simeq id_{T_2}\).
Step 1:
There is a continuous \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\), by Note for the definition of homotopy equivalence.
There is a continuous \(\widetilde{\widetilde{f_3}}: T_3 \to T_1\) such that \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\), likewise.
Step 2:
Let us take \(f_1 \circ \widetilde{\widetilde{f_3}}: T_3 \to T_2\).
\(f_1 \circ \widetilde{\widetilde{f_3}}\) 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.
\(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\), which has been seen above.
\(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 = f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \circ id_{T_2} \simeq f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \circ \widetilde{f_2}\), because \(id_{T_2} \simeq f_1 \circ \widetilde{f_2}\), 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, \(= f_1 \circ (\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1) \circ \widetilde{f_2} \simeq f_1 \circ id_{T_1} \circ \widetilde{f_2}\), because \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\), as before, \(= f_1 \circ \widetilde{f_2} \simeq id_{T_2}\).
By the proposition that on the set of the continuous maps between any topological spaces, being homotopic is an equivalence relation, \(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \simeq id_{T_2}\).
So, \(f_2\) is a homotopy equivalence.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for continuous map from 1st space into 2nd space and continuous map from 2nd space into 3rd space, if 2nd map and composition of 2nd map after 1st map are homotopy equivalences, 1st map is homotopy equivalence
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 for any continuous map from any 1st topological space into any 2nd topological space and any continuous map from the 2nd space into any 3rd topological space, if the 2nd map and the composition of the 2nd map after the 1st map are some homotopy equivalences, the 1st map 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(f_2\): \(: T_2 \to T_3\), \(\in \{\text{ the continuous maps }\}\)
//
Statements:
\(f_2 \in \{\text{ the homotopy equivalences }\} \land f_2 \circ f_1 \in \{\text{ the homotopy equivalences }\}\)
\(\implies\)
\(f_1 \in \{\text{ the homotopy equivalences }\}\)
//
2: Note
\(f_1\) needs to be presupposed to be continuous, for this proposition, because Proof does not prove that \(f_1\) is continuous, which is a necessity for \(f_1\) to be a homotopy equivalence.
3: Proof
Whole Strategy: Step 1: take \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\) and \(\widetilde{\widetilde{f_3}}: T_3 \to T_1\) such that \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\); Step 2: see that \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \simeq id_{T_2}\).
Step 1:
There is a continuous \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\), by Note for the definition of homotopy equivalence.
There is a continuous \(\widetilde{\widetilde{f_3}}: T_3 \to T_1\) such that \(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\), likewise.
Step 2:
Let us take \(\widetilde{\widetilde{f_3}} \circ f_2: T_2 \to T_1\).
\(\widetilde{\widetilde{f_3}} \circ f_2\) 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.
\(\widetilde{\widetilde{f_3}} \circ f_2 \circ f_1 \simeq id_{T_1}\), which has been seen above.
\(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 = id_{T_2} \circ f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \simeq \widetilde{f_3} \circ f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2\), because \(id_{T_2} \simeq \widetilde{f_3} \circ f_2\), 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, \(= \widetilde{f_3} \circ (f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}}) \circ f_2 \simeq \widetilde{f_3} \circ id_{T_3} \circ f_2\), because \(f_2 \circ f_1 \circ \widetilde{\widetilde{f_3}} \simeq id_{T_3}\), as before, \(= \widetilde{f_3} \circ f_2 \simeq id_{T_2}\).
By the proposition that on the set of the continuous maps between any topological spaces, being homotopic is an equivalence relation, \(f_1 \circ \widetilde{\widetilde{f_3}} \circ f_2 \simeq id_{T_2}\).
So, \(f_1\) is a homotopy equivalence.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the homotopy equivalences from } T_1 \text{ into } T_2\}\)
\(f'_1\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
//
Statements:
\(f_1 \simeq f'_1\)
\(\implies\)
\(f'_1 \in \{\text{ the homotopy equivalences from } T_1 \text{ into } T_2\}\)
//
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.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(2\) continuous maps with same domain and codomain and equivalence relations on domain and codomain, if each class is mapped into class and maps are homotopic relative to subset that contains multi-points classes, induced maps between quotient spaces is homotopic relative to quotient of subset
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 for any \(2\) continuous maps with any same domain and codomain and any equivalence relations on the domain and the codomain, if each equivalence class is mapped into an equivalence class and the maps are homotopic relative to any subset that contains all the multi-points equivalence classes, the induced maps between the quotient spaces is homotopic relative to the quotient of the subset.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(f'\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\sim_1\): \(\in \{\text{ the equivalence relations on } T_1\}\)
\(\sim_2\): \(\in \{\text{ the equivalence relations on } T_2\}\)
\(T_1 / \sim_1\): \(= \text{ the quotient topological space }\)
\(T_2 / \sim_2\): \(= \text{ the quotient topological space }\)
\(S_1\): \(\subseteq T_1\), such that \(\forall t_1 \in T_1 \setminus S_1 ([t_1]_1 = \{t_1\})\)
//
Statements:
(
\(\forall t_1, t'_1 \in T_1 \text{ such that } t_1 \sim_1 t'_1 (f (t_1) \sim_2 f (t'_1) \land f' (t_1) \sim_2 f' (t'_1))\)
\(\land\)
\(f \simeq f' rel S_1\)
)
\(\implies\)
\(\widetilde{f} \simeq \widetilde{f'} rel \{[s_1]_1 \vert s_1 \in S_1\} \subseteq T_1 / \sim_1\) where \(\widetilde{f}: T_1 / \sim_1 \to T_2 / \sim_2, [t_1]_1 \mapsto [f (t_1)]_2\) and \(\widetilde{f'}: T_1 / \sim_1 \to T_2 / \sim_2, [t_1]_1 \mapsto [f' (t_1)]_2\)
//
2: Note
\(\{[s_1]_1 \vert s_1 \in S_1\}\) is called "the quotient of the subset" in Title and Target Context, but it is not really so, because "the quotient of the subset" is \(S_1 / \sim_1\) but \(\{[s_1]_1 \vert s_1 \in S_1\}\) is a subset of \(T_1 / \sim_1\), but anyway, \(\{[s_1]_1 \vert s_1 \in S_1\}\) and \(S_1 / \sim_1\) are canonically 'sets - maps' isomorphic.
A typical case that this proposition applies is an adjunction space, \(T_1 = T_{1, 1} + T_{1, 2}\) and \(T_1 / \sim_1 = T_{1, 2} \cup_g T_{1, 1}\) with \(g: S \to T_{1, 2}\) and \(S_1 = S \cup T_{1, 2}\).
The reason why "\(rel S_1\)" is required is that otherwise, \(F\) would not necessarily induce \(\widetilde{F}\) in Proof.
3: Proof
Whole Strategy: Step 1: see that \(\widetilde{f}\) and \(\widetilde{f'}\) are well-defined; Step 2: take any homotopy between \(f\) and \(f'\) relative to \(S_1\), \(F\); Step 3: see that \(\widetilde{F}: (T_1 / \sim_1) \times I \to T_2 / \sim_2\) is induced from \(F\); Step 4: see that \(\widetilde{F}\) is a homotopy between \(\widetilde{f}\) and \(\widetilde{f'}\) relative to \(\{[s_1]_1 \vert s_1 \in S_1\}\).
Step 1:
Let us see that \(\widetilde{f}\) and \(\widetilde{f'}\) are well-defined.
Let \(t_1, t'_1 \in T_1\) be any such that \([t_1]_1 = [t'_1]_1\).
That means that \(t_1 \sim_1 t'_1\).
By the supposition, \(f (t_1) \sim_2 f (t'_1)\).
That means that \([f (t_1)]_2 = [f (t'_1)]_2\).
So, for each \([t_1]_1 \in T_1 / \sim_1\), \([f (t_1)]_2\) is uniquely determined independent of the choice of \(t_1\).
So, \(\widetilde{f}\) is well-defined.
\(\widetilde{f'}\) is well-defined, likewise.
Step 2:
As \(f \simeq f' rel S_1\), there is a homotopy, \(F: T_1 \times I \to T_2\), such that for each \(t_1 \in T_1\), \(F (t_1, 0) = f (t_1)\) and \(F (t_1, 1) = f' (t_1)\) and for each \(s_1 \in S_1\), for each \(r \in I\), \(F (s_1, r) = f (s_1) = f' (s_1)\).
Step 3:
Let us see that \(\widetilde{F}: (T_1 / \sim_1) \times I \to T_2 / \sim_2, ([t_1]_1, r) \mapsto [F (t_1, r)]_2\) is well-defined.
Let \(t_1, t'_1 \in T_1\) be any such that \([t_1]_1 = [t'_1]_1\).
When \(t_1 = t'_1\), \(F (t_1, r) = F (t'_1, r)\), so, \([F (t_1, r)]_2 = [F (t'_1, r)]_2\).
Let us suppose that \(t_1 \neq t'_1\).
\(t_1, t'_1 \in S_1\), because if \(t_1 \in T_1 \setminus S_1\), \([t_1]_1 = \{t_1\}\), by the supposition, so, \(t'_1 \in [t_1]\) would imply that \(t'_1 = t_1\), a contradiction, and likewise for \(t'_1 \in T_1 \setminus S_1\).
So, \(F (t_1, r) = f (t_1) = f' (t_1)\) and \(F (t'_1, r) = f (t'_1) = f' (t'_1)\).
But \([f (t_1)]_2 = [f (t'_1)]_2\), by the supposition.
So, \([F (t_1, r)]_2 = [f (t_1)]_2 = [f (t'_1)]_2 = [F (t'_1, r)]_2\).
So, for each \(([t_1]_1, r) \in (T_1 / \sim_1) \times I\), \([F (t_1, r)]_2\) is uniquely determined independent of the choice of \(t_1\).
So, \(\widetilde{F}\) is well-defined.
Step 4:
\(\widetilde{F}\) is continuous, by the proposition that for any continuous map from the product of any topological space and any locally compact Hausdorff topological space and any equivalence relations on the 1st space and the codomain, if each 1st space equivalence class is mapped into any codomain equivalence class, the induced map between the product of the quotient space and the 2nd space and the quotient space is continuous.
\(\widetilde{F} ([t_1]_1, 0) = [F (t_1, 0)]_2 = [f (t_1)]_2 = \widetilde{f} ([t_1]_1)\) and \(\widetilde{F} ([t_1]_1, 1) = [F (t_1, 1)]_2 = [f' (t_1)]_2 = \widetilde{f'} ([t_1]_1)\).
For each \([s_1]_1 \in \{[s_1]_1 \vert s_1 \in S_1\} \subseteq T_1 / \sim_1\), for each \(r \in I\), \(\widetilde{F} ([s_1]_1, r) = [F (s_1, r)]_2 = [f (s_1)]_2 = [f' (s_1)]_2\), but \([f (s_1)]_2 = \widetilde{f} ([s_1]_1)\) and \([f' (s_1)]_2 = \widetilde{f'} ([s_1]_1)\).
That means that \(\widetilde{F}\) is a homotopy between \(\widetilde{f}\) and \(\widetilde{f'}\) relative to \(\{[s_1]_1 \vert s_1 \in S_1\}\).
So, \(\widetilde{f} \simeq \widetilde{f'} rel \{[s_1]_1 \vert s_1 \in S_1\}\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(2\) continuous maps from same domain into same codomain and disjoint open cover of domain, if restrictions of maps on each element of cover are homotopic relative to subset, maps are homotopic relative to union of subsets
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 for any \(2\) continuous maps from any same domain into any same codomain and any disjoint open cover of the domain, if the restrictions of the maps on each element of the cover are homotopic relative to any subset, the maps are homotopic relative to the union of the subsets.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(f'\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\{U_j \in \{\text{ the open subsets of } T_1\} \vert j \in J\}\): where \(J \in \{\text{ the possibly uncountable index sets }\}\), \(\forall j, j' \in J \text{ such that } j \neq j' (U_j \cap U_{j'} = \emptyset)\), and \(\cup_{j \in J} U_j = T_1\)
//
Statements:
\(\forall j \in J (f \vert_{U_j} \simeq f' \vert_{U_j} rel S_j)\)
\(\implies\)
\(f \simeq f' rel \cup_{j \in J} S_j\)
//
2: Note
Each \(f \vert_{U_j}: U_j \to T_2\) or \(f' \vert_{U_j}: U_j \to T_2\) is inevitably continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, so, talking about being homotopic of \(f \vert_{U_j}\) and \(f' \vert_{U_j}\) makes sense.
\(\{U_j \in \{\text{ the open subsets of } T_1\} \vert j \in J\}\) needs to be disjoint for this proposition, because otherwise, when the homotopy between \(f\) and \(f'\) was constructed from the homotopies between the restricted maps, the consistency would be a concern.
A typical case that this proposition applies is that \(T_1\) is a topological sum, \(\coprod_{j \in J} T_{1, j}\): \(\{T_{1, j}\}\) is a disjoint open cover of \(T_1\).
3: Proof
Whole Strategy: Step 1: for each \(j \in J\), take a homotopy, \(F_j: U_j \times I \to T_2\) between \(f \vert_{U_j}\) and \(f' \vert_{U_j}\) relative to \(S_j\); Step 2: take \(F: T_1 \times I \to T_2\) such that \(F \vert_{U_j \times I} = F_j\), and see that \(F\) is a homotopy between \(f\) and \(f'\) relative to \(\cup_{j \in J} S_j\).
Step 1:
Let \(j \in J\) be any.
As \(f \vert_{U_j} \simeq f' \vert_{U_j} rel S_j\), there is a homotopy, \(F_j: U_j \times I \to T_2\), such that \(F_j (u_j, 0) = f (u_j)\), \(F_j (u_j, 1) = f' (u_j)\), and for each \(s_j \in S_j\), \(F_j (s_j, j) = f (s_j) = f' (s_j)\).
Step 2:
Let us take \(F: T_1 \times I \to T_2\) such that \(F \vert_{U_j \times I} = F_j\).
\(F\) is well-defined, because \(\{U_j \times I \vert j \in J\}\) is a disjoint cover of \(T_1 \times I\).
Each \(U_j \times I\) is an open subset of \(T_1 \times I\), by Note for the definition of product topology.
So, \(\{U_j \times I \vert j \in J\}\) is an open cover of \(T_1 \times I\).
\(F\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous: the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
For each \(t_1 \in T_1\), \(t_1 \in U_j\) for a \(j \in J\), and \(F (t_1, 0) = F_j (t_1, 0) = f (t_1)\).
For each \(t_1 \in T_1\), \(t_1 \in U_j\) for a \(j \in J\), and \(F (t_1, 1) = F_j (t_1, 1) = f' (t_1)\).
For each \(s \in \cup_{j \in J} S_j\), \(s \in S_j \subseteq U_j\) for a \(j \in J\), and for each \(r \in I\), \(F (s, r) = F_j (s, r) = f (s) = f' (s)\).
So, \(F\) is a homotopy between \(f\) and \(f'\) relative to \(\cup_{j \in J} S_j\).
So, \(f \simeq f' rel \cup_{j \in J} S_j\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(2\) continuous maps from same domain into same codomain and finite disjoint closed cover of domain, if restrictions of maps on each element of cover are homotopic relative to subset, maps are homotopic relative to union of subsets
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 for any \(2\) continuous maps from any same domain into any same codomain and any finite disjoint closed cover of the domain, if the restrictions of the maps on each element of the cover are homotopic relative to any subset, the maps are homotopic relative to the union of the subsets.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(f'\): \(\in T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\{C_j \in \{\text{ the closed subsets of } T_1\} \vert j \in J\}\): where \(J \in \{\text{ the finite index sets }\}\), \(\forall j, j' \in J \text{ such that } j \neq j' (C_j \cap C_{j'} = \emptyset)\), and \(\cup_{j \in J} C_j = T_1\)
//
Statements:
\(\forall j \in J (f \vert_{C_j} \simeq f' \vert_{C_j} rel S_j)\)
\(\implies\)
\(f \simeq f' rel \cup_{j \in J} S_j\)
//
2: Note
Each \(f \vert_{C_j}: C_j \to T_2\) or \(f' \vert_{C_j}: C_j \to T_2\) is inevitably continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, so, talking about being homotopic of \(f \vert_{C_j}\) and \(f' \vert_{C_j}\) makes sense.
\(\{C_j \in \{\text{ the closed subsets of } T_1\} \vert j \in J\}\) needs to be disjoint for this proposition, because otherwise, when the homotopy between \(f\) and \(f'\) was constructed from the homotopies between the restricted maps, the consistency would be a concern.
A typical case that this proposition applies is that \(T_1\) is a topological sum, \(T_{1, 1} + ... + T_{1, n}\): \(\{T_{1, j}\}\) is a finite disjoint closed cover of \(T_1\).
3: Proof
Whole Strategy: Step 1: for each \(j \in J\), take a homotopy, \(F_j: C_j \times I \to T_2\) between \(f \vert_{C_j}\) and \(f' \vert_{C_j}\) relative to \(S_j\); Step 2: take \(F: T_1 \times I \to T_2\) such that \(F \vert_{C_j \times I} = F_j\), and see that \(F\) is a homotopy between \(f\) and \(f'\) relative to \(\cup_{j \in J} S_j\).
Step 1:
Let \(j \in J\) be any.
As \(f \vert_{C_j} \simeq f' \vert_{C_j} rel S_j\), there is a homotopy, \(F_j: C_j \times I \to T_2\), such that \(F_j (c_j, 0) = f (c_j)\), \(F_j (c_j, 1) = f' (c_j)\), and for each \(s_j \in S_j\), \(F_j (s_j, j) = f (s_j) = f' (s_j)\).
Step 2:
Let us take \(F: T_1 \times I \to T_2\) such that \(F \vert_{C_j \times I} = F_j\).
\(F\) is well-defined, because \(\{C_j \times I \vert j \in J\}\) is a disjoint cover of \(T_1 \times I\).
Each \(C_j \times I\) is a closed subset of \(T_1 \times I\), by the proposition that for any product topological space, the product of any closed subsets is closed.
So, \(\{C_j \times I \vert j \in J\}\) is a closed cover of \(T_1 \times I\).
\(F\) 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: the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
For each \(t_1 \in T_1\), \(t_1 \in C_j\) for a \(j \in J\), and \(F (t_1, 0) = F_j (t_1, 0) = f (t_1)\).
For each \(t_1 \in T_1\), \(t_1 \in C_j\) for a \(j \in J\), and \(F (t_1, 1) = F_j (t_1, 1) = f' (t_1)\).
For each \(s \in \cup_{j \in J} S_j\), \(s \in S_j \subseteq C_j\) for a \(j \in J\), and for each \(r \in I\), \(F (s, r) = F_j (s, r) = f (s) = f' (s)\).
So, \(F\) is a homotopy between \(f\) and \(f'\) relative to \(\cup_{j \in J} S_j\).
So, \(f \simeq f' rel \cup_{j \in J} S_j\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
