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A definition of topological sum
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of topological sum
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 family of topological spaces, \(\{T_\alpha\}\) where \(\{\alpha\}\) is any possibly uncountable indexes set, the disjoint union, \(\coprod_\alpha T_\alpha\), with the topology such that any subset, \(S \subseteq \coprod_\alpha T_\alpha\) is open if and only if \(S \cap T_\alpha\) is open for each \(\alpha\), often denoted like \(T_1 + T_2\) when the indexes set is finite
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that if union of disjoint subsets is open, each subset is not necessarily open
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 if the union of some disjoint subsets is open, each subset is not necessarily open.
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, \(T\), and some disjoint subsets, \(\{S_i| S_i \subseteq T\}\), such that the union, \(S := \cup_i S_i\), is open, each \(S_i\) is not necessarily open.
2: Proof
A counterexample will suffice. Think of the \(\mathbb{R}\) Euclidean topological space, and 2 subsets, \((-1, 0), [0, 1)\), which are disjoint and the union is \((-1, 1)\), open.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that map preimages of disjoint subsets are disjoint
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 the preimages of any disjoint subsets under any map are disjoint.
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 sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any disjoint subsets, \(S_{21}, S_{22} \subseteq S_2\), such that \(S_{21} \cap S_{22} = \emptyset\), \(f^{-1} (S_{21}) \cap f^{-1} (S_{22}) = \emptyset\).
2: Proof
Suppose that there was a common element, \(p \in f^{-1} (S_{21})\) and \(p \in f^{-1} (S_{22})\). \(f (p) \in S_{21}\) and \(f (p) \in S_{22}\), a contradiction.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that subset of quotient topological space is closed iff preimage of subset under quotient map is closed
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 quotient topological space, any subset is closed if and only if the preimage of the subset under the quotient map is closed.
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, \(T_1\), and any quotient topological space, \(T_2\), with respect to any quotient map, \(f: T_1 \rightarrow T_2\), any subset, \(C \subseteq T_2\), is closed if and only if \(f^{-1} (C)\) is closed.
2: Proof
Suppose that \(f^{-1} (C)\) is closed. \(f^{-1} (T_2 \setminus C) = T_1 \setminus f^{-1} (C)\) by the proposition that the preimage of the codomain minus any codomain subset of any map is the domain minus the preimage of the subset. \(T_1 \setminus f^{-1} (C)\) is open, so, \(T_2 \setminus C\) is open by the definition of quotient map, so, \(C\) is closed.
Suppose that \(C\) is closed. \(T_2 \setminus C\) is open. \(f^{-1} (T_2 \setminus C) = T_1 \setminus f^{-1} (C)\) as before, and is open as the preimage of an open set under a continuous map, so, \(f^{-1} (C)\) is closed.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that map of quotient topology is quotient map
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 map of any quotient topology is a quotient map.
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, \(T\), any set, \(S\), any surjection, \(f: T \rightarrow S\), and the quotient topology on \(S\) with respect to \(f\), \(O\), \(f\) is a quotient map.
2: Proof
\(f\) is obviously a continuous surjection by the definition of quotient topology. For any subset, \(U \subseteq S\), if \(f^{-1} (U)\) is open, \(U\) is open by the definition of quotient topology.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that quotient topology is sole finest topology that makes map continuous
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 topological space, any set, and any surjection from the topological space to the set, the quotient topology is the sole finest topology that makes the map continuous.
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, \(T\), any set, \(S\), and any map, \(f: T \rightarrow S\), the quotient topology on \(S\) with respect to \(f\), \(O\), is the finest topology that makes \(f\) continuous, which means not only that \(O\) is a finest topology but also that any topology that makes \(f\) continuous is coarser than \(O\).
2: Proof
Any topology, \(O'\), that does not equal \(O\) but is not coarser than \(O\) has an open set, \(U\), that is not in \(O\). By the definition of \(O\), \(f^{-1} (U)\) is not open. So, \(f\) is not continuous.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A definition of quotient topology on set with respect to map
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of quotient topology on set with respect to map.
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, \(T\), any set, \(S\), and any surjection, \(f: T \rightarrow S\), the topology on \(S\) defined such that any subset, \(U \subseteq S\), is open if and only \(f^{-1} (U)\) is open
2: Note
The quotient topology is indeed a topology, because \(f^{-1} (\emptyset) = \emptyset\), so, \(\emptyset \subseteq S\) is open; \(f^{-1} (S) = T\), so, \(S \subseteq S\) is open; for any possibly uncountable number of open sets, \(\{U_\alpha\}\), \(f^{-1} (\cup_\alpha U_\alpha) = \cup_\alpha f^{-1} (U_\alpha)\) by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets, which is open; for any finite number of open sets, \(\{U_i\}\), \(f^{-1} (\cap_i U_i) = \cap_i (f^{-1} (U_i))\) by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets, which is open.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A definition of adjunction topological space obtained by attaching topological space via map to topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of adjunction topological space obtained by attaching topological space via map to 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 spaces, \(T_1, T_2\), any subset, \(S \subseteq T_1\), and any continuous map, \(f: S \rightarrow T_2\), the quotient space of the topological sum, \(T_1 + T_2\), such that any point, \(p \in f (S)\), and all the points of the set, \(f^{-1} (p) \subseteq S\), are identified, is the adjunction topological space obtained by attaching \(T_1\) via \(f\) to \(T_2\), denoted as \(T_2 \cup_f T_1\)
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that when image of point is on image of subset, point is on subset if map is injective with respect to image of 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 map between any sets, when the image of any point is on the image of any subset, the point is on the subset if the map is injective with respect to the image 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: Description
For any sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), any point, \(p \in S_1\), and any subset, \(S \subseteq S_1\), when \(f (p) \in f (S)\), \(p \in S\) if \(f\) is injective with respect to \(f (S)\), which means that for any \(q \in f (S)\), \(f^{-1} (q)\) is a 1 element set.
2: Proof
As \(f (p) \in f (S)\), there is a \(p' \in S\) such that \(f (p') = f(p)\). As \(f^{-1} (f (p))\) is a 1 element set, \(p' = p\), so, \(p \in S\).
3: Note
Of course, \(f\) can be wholly injective in order to satisfy the condition.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of universal property of quotient map
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 universal property of quotient map: any surjection between topological spaces is a quotient map if and only if any additional map from the codomain of the original map to any additional topological space is continuous if and only if the composition of the additional map after the original map is continuous.
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, \(T_1, T_2\), and any surjection, \(f_1: T_1 \rightarrow T_2\), \(f_1\) is a quotient map if and only if for any topological space, \(T_3\), and any map, \(f_2: T_2 \rightarrow T_3\), \(f_2\) is continuous if and only if \(f_2 \circ f_1: T_1 \rightarrow T_3\) is continuous.
2: Proof
Suppose that \(f_1\) is a quotient map. Suppose \(f_2\) is continuous. Then, \(f_2 \circ f_1\) is continuous as a compound of continuous maps. Suppose \(f_2 \circ f_1\) is continuous. Then, for any open set, \(U \subseteq T_3\), \({f_2 \circ f_1}^{-1} (U) = {f_1}^{-1} \circ {f_2}^{-1} (U) = {f_1}^{-1} ({f_2}^{-1} (U))\) is open. By the definition of quotient map, \({f_2}^{-1} (U)\) is open, so, \(f_2\) is continuous.
Suppose that for any topological space, \(T_3\), and any map, \(f_2: T_2 \rightarrow T_3\), \(f_2\) is continuous if and only if \(f_2 \circ f_1: T_1 \rightarrow T_3\) is continuous. Let us take \(T_3 = T_2\) and \(f_2: T_2 \rightarrow T_2\) as the identity map, continuous. So, \(f_2 \circ f_1 = f_1: T_1 \rightarrow T_2\) is continuous. Let us take \(T_3 := T_1 / f_1\), which is the quotient space of \(T_1\) such that any pair, \(p_1, p_2 \in T_1, f_1 (p_1) = f_1 (p_2)\), are identified. Let us take \(f_2: f_1 (p) \mapsto [p]\), which is obviously bijective. \(f_2 \circ f_1\), which is really the canonical map to the quotient space, is continuous, because for any open set, \(U \subseteq T_3\), \((f_2 \circ f_1)^{-1} (U)\) is open by the definition of quotient topology. So, by the supposition, \(f_2\) is continuous. Now, for any subset, \(S \subseteq T_2\), if \({f_1}^{-1} (S)\) is open, \(f_2 \circ f_1 ({f_1}^{-1} (S))\) is open by the definition of quotient topology, because \({f_1}^{-1} (S) = (f_2 \circ f_1)^{-1} (f_2 \circ f_1 ({f_1}^{-1} (S))\), which is because for any \(p \in {f_1}^{-1} (S)\), \(f_2 \circ f_1 (p) \in f_2 \circ f_1 ({f_1}^{-1} (S))\), so, \(p \in (f_2 \circ f_1)^{-1} (f_2 \circ f_1 ({f_1}^{-1} (S))\), and for any \(p \in (f_2 \circ f_1)^{-1} (f_2 \circ f_1 ({f_1}^{-1} (S))\), \(f_2 \circ f_1 (p) \in f_2 \circ f_1 ({f_1}^{-1} (S))\), but as \(f_2\) is bijective, \(f_1 (p) \in f_1 ({f_1}^{-1} (S))\), but as \(f_1\) is surjective, by the proposition that for any map, the composition of the map after any preimage is identical if and only if the argument set is a subset of the map image, \(f_1 \circ {f_1}^{-1} (S) = S\), so, \(f_1 (p) \in S\). \(S = {f_2}^{-1} ((f_2 \circ f_1 ({f_1}^{-1} (S)))\) because \(f_2\) is bijective, but as \(f_2\) is continuous, \(S\) is open.
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>
A description/proof of universal property of continuous embedding
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 universal property of continuous embedding: any injection between topological spaces is a continuous embedding if and only if any additional map from any additional topological space into the domain of the original map is continuous if and only if the composition of the additional map before the original map is continuous.
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, \(T_1, T_2\), and any injection, \(f_1: T_1 \rightarrow T_2\), \(f_1\) is a continuous embedding if and only if for any topological space, \(T_3\), and any map, \(f_2: T_3 \rightarrow T_1\), \(f_2\) is continuous if and only if \(f_1 \circ f_2: T_3 \rightarrow T_2\) is continuous.
2: Proof
Suppose that \(f_1\) is a continuous embedding. Suppose \(f_2\) is continuous. Then, \(f_1 \circ f_2\) is continuous as a compound of continuous maps. Suppose \(f_1 \circ f_2\) is continuous. Then, for any open set, \(U \subseteq T_1\), \(f_1 (U)\) is open on \(f_1 (T_1)\) while \(f_1 (U) = U' \cap f_1 (T_1)\) where \(U' \subseteq T_2\) is open on \(T_2\) by the definition of subspace topology. As \(f_1 \circ f_2\) is continuous, \((f_1 \circ f_2)^{-1} (U') = f_2^{-1} \circ f_1^{-1} (U')\) is open on \(T_3\). But \(f_1^{-1} (U') = f_1^{-1} (U' \cap f_1 (T_1)) = f_1^{-1} (f_1 (U))\), but \(f_1^{-1} (f_1 (U)) = U\), because \(f_1\) is bijective. So, \(f_2^{-1} (U)\) is open on \(T_3\), which means that \(f_2\) is continuous.
Suppose that for any topological space, \(T_3\), and any map, \(f_2: T_3 \rightarrow T_1\), \(f_2\) is continuous if and only if \(f_1 \circ f_2: T_3 \rightarrow T_2\) is continuous. Let us take \(T_3 = T_1\) and \(f_2: T_1 \rightarrow T_1\) as the identity map, continuous. So, \(f_1 \circ f_2 = f_1: T_1 \rightarrow T_2\) is continuous. As \(f_1\) is injective, \(f'_1: T_1 \rightarrow f_1 (T_1)\) is bijective, so, let us take \(T_3 := f_1 (T_1)\) where \(T_3\) is the subspace topological space and \(f_2 := {f'_1}^{-1}: f_1 (T_1) \rightarrow T_1\). \(f_1 \circ f_2: f_1 (T_1) \rightarrow T_2\) is an inclusion, so, continuous. So, \(f_2 = {f'_1}^{-1}\) is continuous, so, \(f'_1\) is a homeomorphism.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A definition of quotient map
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of quotient map.
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 spaces, \(T_1, T_2\), any continuous surjection, \(f: T_1 \rightarrow T_2\), such that for any subset, \(S \subseteq T_2\), if \(f^{-1} (S)\) is open, \(S\) is open
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that restriction of continuous map on domain and codomain is continuous
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 restriction of any continuous map on the domain and the codomain is continuous.
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, \(f: T_1 \rightarrow T_2\), any subset, \(S_1 \subseteq T_1\), and any subset, \(S_2 \subseteq T_2\) such that \(f (S_1) \subseteq S_2\), \(f|_{S_1}: S_1 \rightarrow S_2\) is continuous.
2: Proof
For any open set, \(U \subseteq S_2\), \(U = U' \cap S_2\) where \(U' \subseteq T_2\) is open on \(T_2\), by the definition of subspace topology. \({f|_{S_1}}^{-1} (U) = {f|_{S_1}}^{-1} (U' \cap S_2) = f^{-1} (U') \cap S_1\), because for any \(p \in {f|_{S_1}}^{-1} (U' \cap S_2)\), \(f|_{S_1} (p) \in U' \cap S_2\) by the definition of preimage, \(f (p) \in U' \cap S_2 \subseteq U'\), so, \(p \in f^{-1} (U')\), while of course \(p \in S_1\) because \(S_1\) is the domain of \({f|_{S_1}}^{-1}\); for any \(p \in f^{-1} (U') \cap S_1\), \(f|_{S_1} (p) \in U'\) because as \(p\) is on \(S_1\), \(f|_{S_1}\) can operate on it with the same result, which is on \(U'\), but as \(f|_{S_1} (S_1) = f (S_1) \subseteq S_2\), \(f|_{S_1} (p) \in U' \cap S_2\), so, \(p \in {f|_{S_1}}^{-1} (U' \cap S_2)\). As \(f\) is continuous, \(f^{-1} (U')\) is open on \(T_1\), and \(f^{-1} (U') \cap S_1\) is open on \(S_1\) by the definition of subspace topology.
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>
A description/proof of that composition of preimage after map of subset is identical iff it is contained in argument set
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 map, the composition of the preimage after the map of any subset is identical if and only if it is contained in the argument 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: Description
For any sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subset, \(S_3 \subseteq S_1\), \(f^{-1} \circ f (S_3) = S_3\) if and only if \(f^{-1} \circ f (S_3) \subseteq S_3\).
2: Proof
Suppose that \(f^{-1} \circ f (S_3) \subseteq S_3\). For any \(p \in f^{-1} \circ f (S_3)\), \(p \in S_3\). For any \(p \in S_3\), \(f (p) \in f (S_3)\), which means that \(p \in f^{-1} \circ f (S_3)\) by the definition of preimage.
Suppose that \(f^{-1} \circ f (S_3) = S_3\). \(f^{-1} \circ f (S_3) \subseteq S_3\).
3: Note
It is important not to carelessly conclude that \(f^{-1} \circ f (S_3) = S_3\) without checking the condition.
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>
A description/proof of that composition of preimage after map of subset is identical if map is injective with respect to argument set image
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 map, the composition of the preimage after the map of any subset is identical if the map is injective with respect to the argument set image.
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 sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subset, \(S_3 \subseteq S_1\), \(f^{-1} \circ f (S_3) = S_3\) if \(f\) is injective with respect to \(f (S_3)\), which means that for any \(p \in f (S_3)\), \(f^{-1} (p)\) is a 1 element set.
2: Proof
Suppose that \(f\) is injective with respect to \(f (S_3)\). For any \(p \in f^{-1} \circ f (S_3)\), \(f (p) \in f (S_3)\) by the definition of preimage, which means that there is a \(p' \in S_3\) such that \(f (p') = f (p)\), but as \(f^{-1} (f (p))\) is a 1 element set, \(p' = p\), so, \(p \in S_3\). For any \(p \in S_3\), \(f (p) \in f (S_3)\), which means that \(p \in f^{-1} \circ f (S_3)\) by the definition of preimage.
3: Note
It is important not to carelessly conclude that \(f^{-1} \circ f (S_3) = S_3\) without checking the condition.
The condition is not any necessary condition; in fact, the issue is not really being injective, but \(f^{-1} \circ f (S_3) \subseteq S_3\), which is a necessary condition; being injective is just a typical case of guaranteeing \(f^{-1} \circ f (S_3) \subseteq S_3\).
The condition is 'with respect to \(f (S_3)\)', not "with respect to \(S_3\)": the condition is not about the injectivity of \(f|_{S_3}: S_3 \rightarrow S_2\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that composition of map after preimage is identical iff argument set is subset of map image
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 map, the composition of the map after any preimage is identical if and only if the argument set is a subset of the map image.
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 sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subset, \(S_3 \subseteq S_2\), \(f \circ f^{-1} (S_3) = S_3\) if and only if \(S_3 \subseteq f (S_1)\).
2: Proof
Suppose that \(S_3 \subseteq f (S_1)\). For any \(p \in f \circ f^{-1} (S_3)\), \(p \in S_3\) by the definition of preimage. For any \(p \in S_3\), as \(S_3 \subseteq f (S_1)\), \(f^{-1} (p) \neq \emptyset\), and \(f (f^{-1} (p)) = p\) by the definition of preimage, so, \(p \in f \circ f^{-1} (S_3)\).
Suppose that \(f \circ f^{-1} (S_3) = S_3\). Suppose that it was not that \(S_3 \subseteq f (S_1)\). There would be a \(p \in S_3\) such that \(p \notin f (S_1)\). \(f^{-1} (p) = \emptyset\), which would mean that \(p \notin f \circ f^{-1} (S_3)\), so, \(p \notin S_3\), a contradiction.
3: Note
It is important not to carelessly conclude that \(f \circ f^{-1} (S_3) = S_3\) without checking the condition.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
A description/proof of that if preimage of closed set under topological spaces map is closed, map is continuous
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 if the preimage of any closed set under a topological spaces map is closed, the map is continuous.
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, \(T_1, T_2\), and any map, \(f: T_1 \rightarrow T_2\), if for any close set, \(C \subseteq T_2\), \(f^{-1} (C)\) is closed, then \(f\) is continuous.
2: Proof
Suppose that for any close set, \(C \subseteq T_2\), \(f^{-1} (C)\) is closed. For any open set, \(U \subseteq T_2\), \(T_2 \setminus U\) is closed. \(f^{-1} (T_2 \setminus U)\) is closed. \(f^{-1} (U) = T_1 \setminus f^{-1} (T_2 \setminus U)\), open. So, as the preimage of any open set is open, \(f\) is continuous.
References
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A description/proof of that map between topological spaces is continuous if domain restriction of map to each closed set of finite closed cover is continuous
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 map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
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, \(T_1, T_2\), and any map, \(f: T_1 \rightarrow T_2\), if there is a finite closed cover of \(T_1\), \(\{C_i \subseteq T_1\}, \cup_{i} C_i = T_1\), such that each \(f|_{C_i}: C_i \rightarrow T_2\) is continuous, \(f\) is continuous.
2: Proof
For any closed set, \(C \subseteq T_2\), \({f|_{C_i}}^{-1} (C)\) is closed on \(C_i\), and on \(T_1\), by the proposition that any closed set on any closed topological subspace is closed on the base space. \(f^{-1} (C) = \cup_i {f|_{C_i}}^{-1} (C)\), because for any \(p \in f^{-1} (C)\), \(f (p) \in C\), but \(p \in \cup_i C_i\), so, \(f|_{C_i} (p) \in C\) for an \(i\), \(p \in {f|_{C_i}}^{-1} (C)\); for any \(p \in \cup_i {f|_{C_i}}^{-1} (C)\), \(p \in {f|_{C_i}}^{-1} (C)\) for an \(i\), so, \(f|_{C_i} (p) \in C\), so, \(f (p) \in C\), so, \(p \in f^{-1} (C)\). So, \(f^{-1} (C)\) is close as the finite union of closed sets. By the proposition that if the preimage of any closed set under a topological spaces map is closed, the map is continuous, \(f\) is continuous.
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>
A description/proof of that map between topological spaces is continuous if domain restriction of map to each open set of open cover is continuous
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 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.
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, \(T_1, T_2\), and any map, \(f: T_1 \rightarrow T_2\), if there is an open cover of \(T_1\), \(\{U_\alpha \subseteq T_1\}, \cup_{\alpha} U_\alpha = T_1\), such that each \(f|_{U_\alpha}: U_\alpha \rightarrow T_2\) is continuous, \(f\) is continuous.
2: Proof
For any open set, \(U \subseteq T_2\), \({f|_{U_\alpha}}^{-1} (U)\) is open on \(U_\alpha\), and on \(T_1\), by the proposition that any open set on any open subspace topological space is open on the base topological space. \(f^{-1} (U) = \cup_\alpha {f|_{U_\alpha}}^{-1} (U)\), because for any \(p \in f^{-1} (U)\), \(f (p) \in U\), but \(p \in \cup_\alpha U_\alpha\), so, \(f|_{U_\alpha} (p) \in U\) for an \(\alpha\), \(p \in {f|_{U_\alpha}}^{-1} (U)\); for any \(p \in \cup_\alpha {f|_{U_\alpha}}^{-1} (U)\), \(p \in {f|_{U_\alpha}}^{-1} (U)\) for an \(\alpha\), so, \(f|_{U_\alpha} (p) \in U\), so, \(f (p) \in U\), so, \(p \in f^{-1} (U)\). So, \(f^{-1} (U)\) is open as the union of open sets.
References
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