description/proof of that for quotient map from 1st topological space onto 2nd topological space and identity map over 3rd locally compact Hausdorff topological space, product of map and identity map is quotient
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quotient map.
- The reader knows a definition of locally compact topological space.
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of product topological space.
- The reader knows a definition of product map.
- The reader admits 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.
- The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
- The reader admits the proposition that for any locally compact Hausdorff topological space, in any neighborhood around any point, there is an open neighborhood of the point whose (the open neighborhood's) closure is compact and contained in the former neighborhood.
- The reader admits the proposition that for any map from the product of any \(2\) sets into any set, any subset of the 2nd set, and any subset of the 3rd set, the complement of the subset of the 1st set such that the image of the product of each point and the subset of the 2nd set is contained in the subset of the 3rd set is the projection of the intersection of the preimage of the complement of the subset of the 3rd set and the product of the 1st set and the subset of the 2nd set.
- The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
- The reader admits 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.
- The reader admits the proposition that for any product topological space and any constituent such that the other constituents are compact, the projection onto the constituent is closed.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any quotient map from any 1st topological space onto any 2nd topological space and the identity map over any 3rd locally compact Hausdorff topological space, the product of the map and the identity map is quotient.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the quotient maps }\}\)
\(id\): \(: T_3 \to T_3, t_3 \mapsto t_3\)
\(T_1 \times T_3\): \(= \text{ the product topological space }\)
\(T_2 \times T_3\): \(= \text{ the product topological space }\)
\(f \times id\): \(: T_1 \times T_3 \to T_2 \times T_3, (t_1, t_3) \mapsto (f (t_1), t_3)\)
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Statements:
\(f \times id \in \{\text{ the quotient maps }\}\)
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2: Proof
Whole Strategy: apply 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; Step 1: see that \(f \times id\) is a continuous surjection, let \(g: T_2 \times T_3 \to T_4\) be any map, and see that if \(g\) is continuous, \(g \circ (f \times id)\) is continuous; Step 2: see that if \(g \circ (f \times id)\) is continuous, \(g\) is continuous; Step 3: conclude the proposition.
Step 1:
\(f \times id\) is a surjection, because for each \((t_2, t_3) \in T_2 \times T_3\), there is a \(t_1 \in T_1\) such that \(f (t_1) = t_2\), because \(f\) is a surjection, so, \(f \times id (t_1, t_3) = (t_2, t_3)\).
\(f \times id\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
Let \(g: T_2 \times T_3 \to T_4\) be any map where \(T_4\) is any topological space.
If \(g\) is continuous, \(g \circ (f \times id)\) 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.
Step 2:
Let us suppose that \(g \circ (f \times id)\) is continuous.
Let us see that \(g\) is continuous at each point on \(T_2 \times T_3\).
Let \((t_2, t_3) \in T_2 \times T_3\) be any.
Let \(U_{g (t_2, t_3)} \subseteq T_4\) be any open neighborhood of \(g (t_2, t_3)\).
There is a \(t_1 \in T_1\) such that \(f (t_1) = t_2\), because \(f\) is a surjection.
\(g \circ (f \times id) (t_1, t_3) = g (f (t_1), t_3) = g (t_2, t_3)\).
As \(g \circ (f \times id)\) is continuous, there is an open neighborhood of \((t_1, t_3)\), \(U_{(t_1, t_3)} \subseteq T_1 \times T_3\), such that \(g \circ (f \times id) (U_{(t_1, t_3)}) \subseteq U_{g (t_2, t_3)}\), but there are an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T_1\), and an open neighborhood of \(t_3\), \(U_{t_3} \subseteq T_3\), such that \(U_{t_1} \times U_{t_3} \subseteq U_{(t_1, t_3)}\), by the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
So, \(g \circ (f \times id) (U_{t_1} \times U_{t_3}) \subseteq U_{g (t_2, t_3)}\).
There is an open neighborhood of \(t_3\), \({U_{t_3}}^` \subseteq T_3\), such that \(\overline{{U_{t_3}}^`} \subseteq T_3\) is compact and \(\overline{{U_{t_3}}^`} \subseteq U_{t_3}\), by the proposition that for any locally compact Hausdorff topological space, in any neighborhood around any point, there is an open neighborhood of the point whose (the open neighborhood's) closure is compact and contained in the former neighborhood, so, \(g \circ (f \times id) (U_{t_1} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\).
Let us define \(S_2 := \{{t_2}' \in T_2 \vert g (\{{t_2}'\} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\}\).
\(t_2 \in S_2\), because \(g \circ (f \times id) (\{t_1\} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\) while \(g \circ (f \times id) (\{t_1\} \times \overline{{U_{t_3}}^`}) = g (\{f (t_1)\} \times \overline{{U_{t_3}}^`}) = g (\{t_2\} \times \overline{{U_{t_3}}^`})\).
\(f^{-1} (S_2) = \{{t_1}' \in T_1 \vert g \circ (f \times id) (\{{t_1}'\} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\}\), because while \(g \circ (f \times id) (\{{t_1}'\} \times \overline{{U_{t_3}}^`}) = g (\{f ({t_1}')\} \times \overline{{U_{t_3}}^`})\), for each \(p \in f^{-1} (S_2)\), \(f (p) \in S_2\), \(g (\{f (p)\} \times \overline{{U_{t_3}}^`})) \subseteq U_{g (t_2, t_3)}\), which implies that \(p \in \{{t_1}' \in T_1 \vert g \circ (f \times id) (\{{t_1}'\} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\}\); for each \(p \in \{{t_1}' \in T_1 \vert g \circ (f \times id) (\{{t_1}'\} \times \overline{{U_{t_3}}^`}) \subseteq U_{g (t_2, t_3)}\}\), \(g (\{f (p)\} \times \overline{{U_{t_3}}^`})) \subseteq U_{g (t_2, t_3)}\), which implies that \(f (p) \in S_2\), so, \(p \in f^{-1} (S_2)\).
\(T_1 \setminus f^{-1} (S_2) = \pi_1 ((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \cap (T_1 \times \overline{{U_{t_3}}^`}))\), by the proposition that for any map from the product of any \(2\) sets into any set, any subset of the 2nd set, and any subset of the 3rd set, the complement of the subset of the 1st set such that the image of the product of each point and the subset of the 2nd set is contained in the subset of the 3rd set is the projection of the intersection of the preimage of the complement of the subset of the 3rd set and the product of the 1st set and the subset of the 2nd set.
\((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \subseteq T_1 \times T_3\) is closed, because \(g \circ (f \times id)\) is continuous and \(T_4 \setminus U_{g (t_2, t_3)} \subseteq T_4\) is closed, by the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed, and \((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \cap (T_1 \times \overline{{U_{t_3}}^`})\) is closed on \(T_1 \times \overline{{U_{t_3}}^`}\), where \(T_1 \times \overline{{U_{t_3}}^`}\) as the product of the topological subspaces is the topological subspace of \(T_1 \times T_3\), by 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.
With \(\widetilde{\pi_1}: T_1 \times \overline{{U_{t_3}}^`} \to T_1 := \pi_1 \vert_{T_1 \times \overline{{U_{t_3}}^`}}\), \(\pi_1 ((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \cap (T_1 \times \overline{{U_{t_3}}^`})) = \widetilde{\pi_1} ((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \cap (T_1 \times \overline{{U_{t_3}}^`}))\).
But \(\widetilde{\pi_1}\) is closed, by the proposition that for any product topological space and any constituent such that the other constituents are compact, the projection onto the constituent is closed: \(\overline{{U_{t_3}}^`}\) as the compact subset is the compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
So, \(T_1 \setminus f^{-1} (S_2) = \pi_1 ((g \circ (f \times id))^{-1} (T_4 \setminus U_{g (t_2, t_3)}) \cap (T_1 \times \overline{{U_{t_3}}^`}))\) is closed on \(T_1\), so, \(f^{-1} (S_2)\) is open on \(T_1\).
So, \(S_2\) is open on \(T_2\), because \(f\) is quotient.
So, \(S_2\) is an open neighborhood of \(t_2\).
\(g (S_2 \times {U_{t_3}}^`) \subseteq U_{g (t_2, t_3)}\), by the definition of \(S_2\): \({U_{t_3}}^` \subseteq \overline{{U_{t_3}}^`}\).
\(S_2 \times {U_{t_3}}^`\) is an open neighborhood of \((t_2, t_3)\).
So, \(g\) is continuous at \((t_2, t_3)\).
So, \(g\) is continuous.
Step 3:
So, by 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, \(f \times id\) is quotient.
