description/proof of that composition of open maps is open 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 open map.
- The reader knows a definition of composition of maps.
Target Context
- The reader will have a description and a proof of the proposition that the composition of any open maps is open 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:
\(\{T'_1, ..., T'_n\}\): \(\subseteq \{\text{ the topological spaces }\}\)
\(\{T_2, ..., T_{n + 1}\}\): \(\subseteq \{\text{ the topological spaces }\}\), such that \(\forall j \in \{2, ..., n\} (T_j = T'_j)\)
\(\{f_1, ..., f_n\}\): \(f_j: T'_j \to T_{j + 1}\), \(\in \{\text{ the open maps }\}\)
\(f_n \circ ... \circ f_1\): \(T'_1 \to T_{n + 1}\), \(= \text{ the composition }\)
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Statements:
\(f_n \circ ... \circ f_1 \in \{\text{ the open maps }\}\)
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2: Note
The condition that \(T_j = T'_j\) is necessarily for this proposition, while 'composition' in general requires only \(T_j \subseteq T'_j\).
For example, let \(T'_1 = \mathbb{R}\), \(T'_2 = \mathbb{R}^2\), \(T_2 = \mathbb{R} \times \{0\}\), and \(T_3 = \mathbb{R}^2\) as the Euclidean topological spaces (strictly speaking, \(T_2\) is the product topological space of Euclidean \(\mathbb{R}\) and \(\{0\}\)) and \(f_1: T'_1 \to T_2, t \mapsto (t, 0)\) and \(f_2: T'_2 \to T_3 = id\), then, \(f_1\) and \(f_2\) are open, but \(f_2 \circ f_1: T'_1 \to T_3\) is not open, because for open \((-1, 1) \subseteq T'_1\), \(f_2 \circ f_1 ((-1, 1)) = (-1, 1) \times \{0\} \subseteq T_3\) is not open.
3: Proof
Whole Strategy: prove it inductively; Step 1: suppose that \(n = 2\), and prove it; Step 2: suppose that the proposition holds for \(n \in \{2, ..., n' - 1\}\) and suppose that \(n = n'\), and prove it; Step 3: conclude the proposition.
Step 1:
Let us suppose that \(n = 2\).
Let \(U \subseteq T'_1\) be any open subset.
\(f_1 (U) \subseteq T_2\) is open, because \(f_1\) is open.
\(f_1 (U)\) is open on \(T'_2\), because \(T'_2 = T_2\).
\(f_2 \circ f_1 (U) = f_2 (f_1 (U)) \subseteq T_3\) is open, because \(f_2\) is open.
So, \(f_2 \circ f_1\) is open.
Step 2:
Let us suppose that the proposition holds for \(n \in \{2, ..., n' - 1\}\) where \(3 \le n'\).
Let us suppose that \(n = n'\).
\(f_n \circ ... \circ f_1 = f_n \circ (f_{n - 1} \circ ... \circ f_1)\).
\(f_{n - 1} \circ ... \circ f_1\) is open, by the induction hypothesis.
\(f_n: T'_n \to T_{n + 1}\) and \(f_{n - 1} \circ ... \circ f_1: T'_1 \to T_n\) where \(T_n = T'_n\), so, the proposition for the case \(n = 2\) applies, and \(f_n \circ ... \circ f_1\) is open.
Step 3:
By the induction principle, the proposition holds for each \(n \in \mathbb{N} \setminus \{0\}\).
