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A description/proof of that for covering map, 2 lifts of continuous map from connected topological space totally agree or totally disagree
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
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The reader will have a description and a proof of the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree.
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: Description
For any connected and locally path-connected topological spaces, \(T_1, T_2\), any covering map, \(\pi: T_1 \to T_2\), which means that \(\pi\) is continuous and surjective and around any point, \(p \in T_2\), there is a neighborhood, \(N_p \subseteq T_2\), that is evenly covered by \(\pi\), any connected topological space, \(T_3\), and any continuous map, \(f: T_3 \to T_2\), any 2 lifts of \(f\), \(f', f'': T_3 \to T_1\), agree at each point of \(T_3\) or disagree at each point of \(T_3\).
2: Proof
The subspace, \(\pi^{-1} (N_p)\), may consist of multiple connected components, each denoted as \({\pi^{-1} (N_p)}_\alpha\) where \(\alpha \in A_p\) where \(A_p\) is a possibly uncountable indices set.
We can take an open neighborhood, \(U_p \subseteq T_2\), as \(N_p\), because if \(N_p\) is not open, there is an open neighborhood, \(U_p \subseteq N_p\), which is homeomorphic to each \({\pi^{-1} (U_p)}_\alpha\) by \(\pi_{p, \alpha} := \pi\vert_{{\pi^{-1} (U_p)}_\alpha}: {\pi^{-1} (U_p)}_\alpha \to U_p\), because \(\pi_{p, \alpha}\) is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of \({\pi^{-1} (N_p)}_\alpha\) and \(N_p\) respectively as a restriction of continuous \(\pi\vert_{{\pi^{-1} (N_p)}_\alpha}: {\pi^{-1} (N_p)}_\alpha \to N_p\), by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, and its inverse is continuous with the domain and the codomain regarded likewise as a restriction of continuous \({\pi\vert_{{\pi^{-1} (N_p)}_\alpha}}^{-1}\), likewise, but by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, those maps are continuous also with the domain and the codomain regarded as subspaces of \(T_1\) and \(T_2\) respectively.
\({\pi^{-1} (U_p)}_\alpha\) is open on \(T_1\), because \(\pi^{-1} (U_p)\) is open on \(T_1\) and is locally path-connected, by the proposition that any open subset of any locally path-connected topological space is locally path-connected, \({\pi^{-1} (U_p)}_\alpha\) is open on \(\pi^{-1} (U_p)\), by the proposition that any connected component on any locally path-connected topological space is open, and \({\pi^{-1} (U_p)}_\alpha\) is open on \(T_1\), by the proposition that any open set on any open topological subspace is open on the base space.
\(\pi \circ f' = \pi \circ f'' = f\) by the definition of lift of continuous map with respect to covering map.
For any point, \(p \in T_3\), there are an \(U_{f (p)} \subseteq T_2\) and \(\{{\pi^{-1} (U_{f (p)})}_\alpha \subseteq T_1\}\). There are 2 possibilities: 1) \(f' (p) = f'' (p) \in {\pi^{-1} (U_{f (p)})}_\alpha\) for an \(\alpha\); 2) \(f' (p) \neq f'' (p)\) where \(f' (p) \in {\pi^{-1} (U_{f (p)})}_{\alpha'}\) and \(f'' (p) \in {\pi^{-1} (U_{f (p)})}_{\alpha''}\) for some \(\alpha' \neq \alpha''\).
Let us suppose the possibility 1). As \({\pi^{-1} (U_{f (p)})}_\alpha\) is open on \(T_1\) and \(f'\) is continuous, there is an open neighborhood, \(U'_p \subseteq T_3\), such that \(f' (U'_p) \subseteq {\pi^{-1} (U_{f (p)})}_\alpha\). Likewise, there is an open neighborhood, \(U''_p \subseteq T_3\), such that \(f'' (U''_p) \subseteq {\pi^{-1} (U_{f (p)})}_\alpha\). Let us define \(U_p := U'_p \cap U''_p\). For any \(p' \in U_p\), \(f' (p'), f'' (p') \in {\pi^{-1} (U_{f (p)})}_\alpha\), \(\pi_{p, \alpha} \circ f' (p') = \pi_{p, \alpha} \circ f'' (p')\), but as \(\pi_{p, \alpha}\) is bijective, \(f' (p') = f'' (p')\). So, for any point, \(p \in T_3\), if \(f' (p) = f'' (p)\), \(f'\) and \(f''\) agree on a neighborhood of \(p\).
Let us suppose the possibility 2). As \({\pi^{-1} (U_{f (p)})}_{\alpha'}\) is open on \(T_1\) and \(f'\) is continuous, there is an open neighborhood, \(U'_p \subseteq T_3\), such that \(f' (U'_p) \subseteq {\pi^{-1} (U_{f (p)})}_{\alpha'}\). Likewise, there is an open neighborhood, \(U''_p \subseteq T_3\), such that \(f'' (U''_p) \subseteq {\pi^{-1} (U_{f (p)})}_{\alpha''}\). Let us define \(U_p := U'_p \cap U''_p\). For any \(p' \in U_p\), \(f' (p') \in {\pi^{-1} (U_{f (p)})}_{\alpha'}\) and \(f'' (p') \in {\pi^{-1} (U_{f (p)})}_{\alpha''}\). As \({\pi^{-1} (U_{f (p)})}_{\alpha'} \cap {\pi^{-1} (U_{f (p)})}_{\alpha''} = \emptyset\), \(f' (p') \neq f'' (p')\). So, for any point, \(p \in T_3\), if \(f' (p) \neq f'' (p)\), \(f'\) and \(f''\) disagree on a neighborhood of \(p\).
So, \(f'\) and \(f''\) agree at each point of \(T_3\) or disagree at each point of \(T_3\), by the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.
References
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A definition of Banach space
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Banach 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: Definition
Any normed vectors space that is a complete metric space with the metric induced by the norm
2: Note
Being complete is about metric space, so, the vectors space has to be given the metric.
References
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A definition of separable topological space
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topological space
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Starting Context
Target Context
-
The reader will have a definition of separable 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: Definition
Any topological space that has a countable dense subset
References
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A definition of nowhere dense subset of topological space
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topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of nowhere dense subset of 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: Definition
Any subset of any topological space whose (the subset's) closure interior is the empty set
References
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A definition of dense subset of topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of dense subset of 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: Definition
Any subset of any topological space whose (the subset's) closure is the topological space
References
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A description/proof of that rotation in \(n\)-dimensional Euclidean vectors space is same \(2\)-dimensional rotations along \((n - 2)\)-dimensional subspace axis
The table of contents of this article
Main Body
References
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A description/proof of that continuous image of path-connected subspace of domain is path-connected on codomain
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 any continuous map image of any path-connected subspace of the domain is path-connected on the codomain.
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: Description
For any topological spaces, \(T_1, T_2\), any continuous map, \(f: T_1 \to T_2\), and any path-connected subspace, \(T_3 \subseteq T_1\), the image, \(f (T_3)\), is path-connected on \(T_2\).
2: Proof
For any points, \(p_1, p_2 \in f (T_3)\), is there a path, \(\lambda: [0, r] \to f (T_3)\) such that \(p_1 = \lambda (0)\) and \(p_2 = \lambda (r)\)? We can take any point from \(f^{-1} (p_i) \cap T_3\) as \(p'_i\). As \(T_3\) is path-connected, there is a path, \(\lambda': [0, 1] \to T_3\), such that \(\lambda' (0) = p'_1\) and \(\lambda' (1) = p'_2\). Let us define \(\lambda:= f \vert_{T_3} \circ \lambda': [0, 1] \to T_3 \to f (T_3)\), which is continuous as a composition of continuous maps (\(f \vert_{T_3}\) is continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous), and \(\lambda (0) = p_1\) and \(\lambda (1) = p_2\).
3: Note
If the domain is a path-connected topological space, the domain is a path-connected subspace of itself, so, the image of any path-connected topological space is path-connected.
References
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A definition of Cauchy sequence on metric space
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metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Cauchy sequence on metric 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: Definition
For any metric space, \(M\), any sequence, \(s: \mathbb{N} \to M\), where \(\mathbb{N}\) is the positive natural numbers set, such that for any \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for any \(n_1, n_2 \in \mathbb{N}\) such that \(N \lt n_1, n_2\), \(dist (s (n_1), s (n_2)) \lt \epsilon\)
2: Note
Sometimes, we may let \(\mathbb{N}\) be the natural numbers set (with \(0\)).
References
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A definition of convergence of sequence on metric space
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metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of convergence of sequence on metric 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: Definition
For any metric space, \(M\), and any sequence, \(s: \mathbb{N} \to M\), where \(\mathbb{N}\) is the positive natural numbers set, a point, \(p \in M\), such that for any \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for any \(n \in \mathbb{N}\) such that \(N \lt n\), \(dist (p, s (n)) \lt \epsilon\)
2: Note
Sometimes, we may let \(\mathbb{N}\) be the natural numbers set (with \(0\)).
When \(M\) is made the topological space with the canonical induced topology, \(\mathbb{N}\) is a directed set, \(s\) is a net with the directed index set, and any convergence of \(s\) as the sequence is a convergence of \(s\) as the net with the directed index set, because for any open neighborhood, \(U_p \subseteq M\), of \(p\), there is an \(\epsilon\) open ball, \(B_{p, \epsilon} \subseteq U_p\), around \(p\), and as \(s (n) \in B_{p, \epsilon}\), \(s (n) \in U_p\).
References
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A definition of product set
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About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of product set.
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: Definition 1
For any possibly uncountable indices set, \(A\), and any sets, \(\{S_\alpha \vert \alpha \in A\}\), indexed with \(A\), the set of all the functions, \(\{f: A \to \cup_{\alpha \in A} S_\alpha \vert f (\alpha) \in S_\alpha \text{ for each } \alpha \in A\}\), denoted as \(\times_{\alpha \in A} S_\alpha\)
2: Definition 2
For any finite number of sets, \(S_1, S_2, ..., S_n\), the set of all the \(n\)-tuples, \(\{\langle p_1, p_2, ..., p_n \rangle \vert p_j \in S_j \text{ for each } j = 1 \sim n\}\), denoted as \(S_1 \times S_2 \times ... \times S_n\)
3: Note
Definition 1 is indeed a set: the set of all the functions from \(A\) into \(\cup_{\alpha \in A} S_\alpha\) is a set (see Proof 8 of the proposition that some expressions can be parts of legitimate formulas for the ZFC set theory), and the formula for the subset axiom is legitimate.
Definition 2 is indeed a set, by the proposition that the product of any finite number of sets is a set.
References
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A description/proof of that subset minus subset is complement of 2nd subset minus complement of 1st subset
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, any subset (the 1st subset) minus any subset (the 2nd subset) is the complement of the 2nd subset minus the complement of the 1st 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: Description
For any set, \(S\), and any subsets, \(S_1, S_2 \subseteq S\), \(S_1 \setminus S_2 = (S \setminus S_2) \setminus (S \setminus S_1)\).
2: Proof
For any element, \(p \in S_1 \setminus S_2\), \(p \in S_1\) and \(p \notin S_2\), \(p \in S \setminus S_2\) and \(p \notin S \setminus S_1\), so, \(p \in (S \setminus S_2) \setminus (S \setminus S_1)\).
For any element, \(p \in (S \setminus S_2) \setminus (S \setminus S_1)\), \(p \in S \setminus S_2\) and \(p \notin S \setminus S_1\), \(p \in S_1\) and \(p \notin S_2\), so, \(p \in S_1 \setminus S_2\).
References
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A description/proof of that union of indexed subsets minus union of subsets indexed with same indices set is contained in union of subset minus subset for each index
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 and any possibly uncountable indices set, the union of any subsets (the 1st set of subsets) indexed with the indices set minus the union of any subsets (the 2nd set of subsets) indexed with the indices set is contained in the union of the subset from the 1st set minus the subset from the 2nd set for each index.
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: Description
For any set, \(S\), any possibly uncountable indices set, \(A\), and any sets of subsets, \(\{S_\alpha\vert \alpha \in A\}\) and \(\{S'_\alpha\vert \alpha \in A\}\), \(\cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha \subseteq \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\).
2: Proof
For any element, \(p \in \cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha\), \(p \in S_\alpha\) for an \(\alpha\) and \(p \notin S'_\alpha\) for each \(\alpha\). \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), \(p \in \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\).
3: Note
Equality does not hold in general, because for any \(p \in \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\), \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), \(p \in S_\alpha\) and \(p \notin S'_\alpha\) for the \(\alpha\), but if \(p \in S'_\beta\) for a \(\beta \neq \alpha\), \(p \notin \cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha\).
References
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A definition of bijection
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of bijection.
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: Definition
For any sets, \(S_1\) and \(S_2\), any map, \(f: S_1 \to S_2\), that is an injection and a surjection
References
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A definition of interior of subset of topological space
Topics
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topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of interior of subset of 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: Definition
For any topological space, \(T\), and any subset, \(S \subseteq T\), the largest open subset, \(int S \subseteq T\), contained in \(S\), which is \(int S \subseteq S\)
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A definition of complete metric space
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of complete metric 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: Definition
Any metric space, \(M\), such that any Cauchy sequence on \(M\) converges to a point on \(M\)
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A description/proof of that 2 continuous maps from connected topological space into Hausdorff topological space such that, for any point, if they agree at point, they agree on neighborhood, totally agree or totally disagree
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 any 2 continuous maps from any connected topological space into any Hausdorff topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood, totally agree or totally disagree on the whole domain.
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: Description
For any connected topological space, \(T_1\), any Hausdorff topological space, \(T_2\), and any continuous maps, \(f_1, f_2: T_1 \to T_2\), such that if \(f_1 (p) = f_2 (p)\) at any point, there is a neighborhood, \(N_p\), of \(p\) such that \(f_1 (p') = f_2 (p')\) for each \(p' \in N_p\), \(f_1 (p) = f_2 (p)\) for each \(p \in T_1\) or \(f_1 (p) \neq f_2 (p)\) for each \(p \in T_1\).
2: Proof
\(f_1\) and \(f_2\) disagree on a neighborhood if they disagree at any point, by the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point. So, \(f_1\) and \(f_2\) totally agree or totally disagree on whole \(T_1\), by the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.
3: Note
Any Euclidean topological space is Hausdorff, so, any 2 continuous maps from any connected topological space into any Euclidean topological space that (the maps) agree on a neighborhood if they agree at any point totally agree or totally disagree on the whole domain.
References
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A description/proof of that 2 continuous maps into Hausdorff topological space that disagree at point disagree on neighborhood of 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 any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the 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: Description
For any topological space, \(T_1\), any Hausdorff topological space, \(T_2\), and any continuous maps, \(f_1, f_2: T_1 \to T_2\), if \(f_1 (p) \neq f_2 (p)\) at any point, \(p \in T_1\), there is a neighborhood, \(N_p\), of \(p\) such that \(f_1 (p') \neq f_2 (p')\) for each point, \(p' \in N_p\).
2: Proof
Let us suppose that \(f_1 (p) \neq f_2 (p)\). There are some disjoint open neighborhoods, \(U_{f_1 (p)}, U_{f_2 (p)} \in T_2\), of \(f_1 (p), f_2 (p)\), respectively, such that \(U_{f_1 (p)} \cap U_{f_2 (p)} = \emptyset\), because \(T_2\) is Hausdorff. As \(f_i\) is continuous, there is a neighborhood, \(N_{p, i}\), of \(p\) such that \(f_i (N_{p, i}) \subseteq U_{f_i (p)}\). \(N_p := N_{p, 1} \cap N_{p, 2}\) is a neighborhood, and \(f_i (N_p) \subseteq U_{f_i (p)}\). \(f_1 (N_p) \cap f_2 (N_p) = \emptyset\), which means that \(f_1 (p') \neq f_2 (p')\) for each \(p' \in N_p\).
References
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