description/proof of that for continuous map and equivalence relations on domain and codomain, if domain equivalence class is mapped into codomain equivalence class, induced map between quotient spaces is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- 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 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 and any equivalence relations on the domain and the codomain, if each domain equivalence class is mapped into any codomain equivalence class, the induced map between the quotient spaces 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\sim_1\): \(\in \{\text{ the equivalence relations on } T_1\}\)
\(\sim_2\): \(\in \{\text{ the equivalence relations on } T_2\}\)
\(T_1 / \sim_1\): \(= \text{ the quotient topological space }\)
\(T_2 / \sim_2\): \(= \text{ the quotient topological space }\)
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Statements:
\(\forall t_1, t'_1 \in T_1 \text{ such that } t_1 \sim_1 t'_1 (f (t_1) \sim_2 f (t'_1))\)
\(\implies\)
\(\widetilde{f}: T_1 / \sim_1 \to T_2 / \sim_2, [t_1]_1 \mapsto [f (t_1)]_2 \in \{\text{ the continuous maps }\}\)
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2: 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 see that \({f_1}^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) = f^{-1} ({f_2}^{-1} (\widetilde{U_2}))\); 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 \(t_1 \sim_1 t'_1\).
\(f (t_1) \sim_2 f (t'_1)\), by the supposition.
That means that \([f (t_1)]_2 = [f (t'_1)]_2\).
So, \([f (t_1)]_2\) is uniquely determined from \([t_1]_1\) 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 us see that \(\widetilde{f} \circ f_1 = f_2 \circ f\).
Let \(t_1 \in T_1\) be any.
\(\widetilde{f} \circ f_1 (t_1) = \widetilde{f} ([t_1]_1) = [f (t_1)]_2\).
\(f_2 \circ f (t_1) = [f (t_1)]_2\).
So, \(\widetilde{f} \circ f_1 (t_1) = f_2 \circ f (t_1)\) for each \(t_1 \in T_1\), which means that \(\widetilde{f} \circ f_1 = f_2 \circ f\).
Let \(\widetilde{U_2} \subseteq T_2 / \sim_2\) be any open subset.
\((\widetilde{f} \circ f_1)^{-1} (\widetilde{U_2}) = (f_2 \circ f)^{-1} (\widetilde{U_2})\).
But the left hand side is \({f_1}^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2}))\) 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, \({f_1}^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) = f^{-1} ({f_2}^{-1} (\widetilde{U_2}))\).
Step 3:
\(f^{-1} ({f_2}^{-1} (\widetilde{U_2})) \subseteq T_1\) is open, because \(f_2\) and \(f\) are continuous.
So, \({f_1}^{-1} ({\widetilde{f}}^{-1} (\widetilde{U_2})) \subseteq T_1\) is open.
As \(f_1\) is quotient, \({\widetilde{f}}^{-1} (\widetilde{U_2}) \subseteq T_1 / \sim_1\) is open.
That means that \(\widetilde{f}\) is continuous.
