description/proof of that for topological space with equivalence relation and subspace with subset equivalence relation, canonical injection from quotient topological space of subspace into quotient topological space of space is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of subset equivalence relation.
- The reader knows a definition of quotient topology on set with respect to map.
- The reader knows a definition of injection.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that for any set with any equivalence relation and any subset with the subset equivalence relation, there is the canonical injection from the quotient set of the subset into the quotient set of the set.
- The reader admits the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.
- 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 topological space with any equivalence relation and any subspace with the subset equivalence relation, the canonical injection from the quotient topological space of the subspace into the quotient topological space of the 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:
\(T'\): \(\in \{\text{ the topological spaces }\}\), with any equivalence relation, \(\sim'\)
\(T\): \(\subseteq T'\), with the subset equivalence relation, \(\sim\)
\(f'\): \(: T' \to T' / \sim', t' \mapsto [t']'\)
\(f\): \(: T \to T / \sim, t \mapsto [t]\)
\(T' / \sim'\): \(= \text{ the quotient set with the quotient topology with respect to } f'\)
\(T / \sim\): \(= \text{ the quotient set with the quotient topology with respect to } f\)
\(g\): \(: T / \sim \to T' / \sim', [t] \mapsto [t]'\)
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Statements:
\(g \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f' \vert_T = g \circ f\); Step 2: see that for each open \(U' \subseteq T' / \sim'\), \({f' \vert_T}^{-1} (U') = (g \circ f)^{-1} (U') = f^{-1} (g^{-1} (U'))\), and see that \(g^{-1} (U')\) is open.
Step 1:
\(g\) is valid and \(f' \vert_T = g \circ f\), by the proposition that for any set with any equivalence relation and any subset with the subset equivalence relation, there is the canonical injection from the quotient set of the subset into the quotient set of the set.
Step 2:
Let \(U' \subseteq T' / \sim'\) be any open subset.
\({f' \vert_T}^{-1} (U') = (g \circ f)^{-1} (U')\).
\({f' \vert_T}^{-1} (U') = f'^{-1} (U') \cap T\), by the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain, which is open on \(T\), because \(f'^{-1} (U') \subseteq T'\) is open, because \(f'\) is continuous with \(T' / \sim'\) given the quotient topology.
\((g \circ f)^{-1} (U') = f^{-1} (g^{-1} (U'))\), 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, \(f^{-1} (g^{-1} (U')) \subseteq T\) is open.
Then, \(g^{-1} (U') \subseteq T / \sim\) is open, by the definition of quotient topology.
So, \(g\) is continuous.
