description/proof of that for continuous map from product of space and locally compact Hausdorff space and equivalence relations on 1st space and codomain, if 1st space equivalence class is mapped into codomain equivalence class, induced map between product of quotient space and 2nd space and quotient space is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- 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 continuous, topological spaces map.
- The reader knows a definition of quotient topology on set with respect to map.
- The reader admits 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.
- The reader admits the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
Target Context
- The reader will have a description and a proof of 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.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(f\): \(: T_1 \times T_3 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\sim_1\): \(\in \{\text{ the equivalence relations on } T_1\}\)
\(\sim_2\): \(\in \{\text{ the equivalence relations on } T_2\}\)
\(T_1 / \sim_1\): \(= \text{ the quotient topological space }\)
\(T_2 / \sim_2\): \(= \text{ the quotient topological space }\)
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Statements:
\(\forall t_1, t'_1 \in T_1 \text{ such that } t_1 \sim_1 t'_1, \forall t_3 \in T_3 (f ((t_1, t_3)) \sim_2 f ((t'_1, t_3)))\)
\(\implies\)
\(\widetilde{f}: (T_1 / \sim_1) \times T_3 \to T_2 / \sim_2, ([t_1]_1, t_3) \mapsto [f ((t_1, t_3))]_2 \in \{\text{ the continuous maps }\}\)
//
2: Note
For example, \(\mathbb{R}\), the Euclidean topological space, is locally compact Hausdorff, by the proposition that any Euclidean metric space is locally compact.
Any open or closed subspace of \(\mathbb{R}\) is locally compact Hausdorff, by the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact, the proposition that any closed subspace of any locally compact topological space is locally compact, and the proposition that any subspace of any Hausdorff topological space is Hausdorff.
Especially, any open or close interval of \(\mathbb{R}\) is locally compact Hausdorff.
Especially, \(I = [0, 1]\) is locally compact Hausdorff.
So, this proposition applies to, for example, \(f: T_1 \times I \to T_2\).
3: Proof
Whole Strategy: Step 1: see that \(\widetilde{f}\) is well-defined; Step 2: take the classification maps, \(f_1: T_1 \to T_1 / \sim_1\) and \(f_2: T_2 \to T_2 / \sim_2\), and \(g: T_1 \times T_3 \to (T_1 / \sim_1) \times T_3 = f_1 \times id\), and see that \(g\) is quotient and \(g^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) = f^{-1} ({f_2}^{-1} (\widetilde{U_2}))\); Step 3: conclude the proposition.
Step 1:
Let us see that \(\widetilde{f}\) is indeed well-defined.
Let \(t_1, t'_1 \in T_1\) be any such that \([t_1]_1 = [t'_1]_1\).
That means that \(t_1 \sim_1 t'_1\).
\(f (t_1, t_3) \sim_2 f (t'_1, t_3)\) for each \(t_3 \in T_3\), by the supposition.
That means that \([f (t_1, t_3)]_2 = [f (t'_1, t_3)]_2\).
So, \([f (t_1, t_3)]_2\) is uniquely determined from \(([t_1]_1, t_3)\) independent of the choice of \(t_1\) in \([t_1]_1\).
So, \(\widetilde{f}\) is well-defined.
Step 2:
Let \(f_1: T_1 \to T_1 / \sim_1\) and \(f_2: T_2 \to T_2 / \sim_2\) be the classification maps.
Let \(g: T_1 \times T_3 \to (T_1 / \sim_1) \times T_3, (t_1, t_3) \mapsto ([t_1], t_3)\).
That is, in fact, \(f_1 \times id\).
\(g\) is a quotient map, by 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.
Let us see that \(\widetilde{f} \circ g = f_2 \circ f\).
Let \((t_1, t_3) \in T_1 \times T_3\) be any.
\(\widetilde{f} \circ g ((t_1, t_3)) = \widetilde{f} (([t_1]_1, t_3)) = [f ((t_1, t_3))]_2\).
\(f_2 \circ f ((t_1, t_3)) = [f ((t_1, t_3))]_2\).
So, \(\widetilde{f} \circ g ((t_1, t_3)) = f_2 \circ f ((t_1, t_3))\) for each \((t_1, t_3) \in T_1 \times T_3\), which means that \(\widetilde{f} \circ g = f_2 \circ f\).
Let \(\widetilde{U_2} \subseteq T_2 / \sim_2\) be any open subset.
\((\widetilde{f} \circ g)^{-1} (\widetilde{U_2}) = (f_2 \circ f)^{-1} (\widetilde{U_2})\).
But the left hand side is \(g^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2}))\) and the right hand side is \(f^{-1} ({f_2}^{-1} (\widetilde{U_2}))\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
So, \(g^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) = f^{-1} ({f_2}^{-1} (\widetilde{U_2}))\).
Step 3:
\(f^{-1} ({f_2}^{-1} (\widetilde{U_2})) \subseteq T_1 \times T_3\) is open, because \(f_2\) and \(f\) are continuous.
So, \(g^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) \subseteq T_1 \times T_3\) is open.
As \(g\) is quotient, \({\widetilde{f}}^{-1} (\widetilde{U_2}) \subseteq (T_1 / \sim_1) \times T_3\) is open.
That means that \(\widetilde{f}\) is continuous.
