description/proof of that finite composition of quotient maps is quotient, if codomains of constituent maps equal domains of succeeding maps
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 composition of maps.
- 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 any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections.
- 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 any finite composition of quotient maps is quotient, if the codomains of the constituent maps equal the domains of the succeeding maps.
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:
\(\{T_1, ..., T_{n + 1}\}\): \(\subseteq \{\text{ the topological spaces }\}\)
\(\{f_1: T_1 \to T_2, ..., f_n: T_n \to T_{n + 1}\}\): \(\subseteq \{\text{ the quotient maps }\}\)
\(f_n \circ ... \circ f_1\): \(: T_1 \to T_{n + 1}\)
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Statements:
\(f_n \circ ... \circ f_1 \in \{\text{ the quotient maps }\}\)
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2: Note
The requirement that "the codomains of the constituent maps equal the domains of the succeeding maps" is imperative, because otherwise, \(f_n \circ ... \circ f_1\) is not even guaranteed to be surjective, by the proposition that a finite composition of surjections is not necessarily any surjection, and no non-surjective map can be quotient.
3: Proof
Whole Strategy: Step 1: see that \(f_n \circ ... \circ f_1\) is a continuous surjection; Step 2: see that for each \(S \subseteq T_{n + 1}\) such that \((f_n \circ ... \circ f_1)^{-1} (S) \subseteq T_1\) is open, \(S\) is open.
Step 1:
\(f_n \circ ... \circ f_1\) 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.
\(f_n \circ ... \circ f_1\) is a surjection, by the proposition that any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections.
Step 2:
Let \(S \subseteq T_{n + 1}\) be any such that \((f_n \circ ... \circ f_1)^{-1} (S) \subseteq T_1\) is open.
\((f_n \circ ... \circ f_1)^{-1} (S) = f_1^{-1} (f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S) \cap S_n) ...) \cap S_2))\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order, \(= f_1^{-1} (f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S)) ...)))\): "\(\cap S_n\)", e.t.c. are unnecessary, because \(S'_j = S_j\) in this case.
As it is open, \(f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S)) ...) \subseteq T_2\) is open, because \(f_1\) is quotient, ..., \(f_n^{-1} (S) \subseteq T_n\) is open, because \(f_{n - 1}\) is quotient, and \(S\) is open, because \(f_n\) is quotient.
So, \(f_n \circ ... \circ f_1\) is quotient.
