description/proof of that for map and its extension that maps extended area outside original codomain, extension preimage of subset is union of original map preimage of intersection of subset and original codomain and extension preimage of subset minus original codomain
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map preimage of subset of codomain.
Target Context
- The reader will have a description and a proof of the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain.
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:
\(S'_1\): \(\in \{\text{ the sets }\}\)
\(S'_2\): \(\in \{\text{ the sets }\}\)
\(S_1\): \(\subseteq S'_1\)
\(S_2\): \(\subseteq S'_2\)
\(f\): \(: S_1 \to S_2\)
\(f'\): \(: S'_1 \to S'_2\), such that \(f' (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\)
\(S\): \(\subseteq S'_2\)
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Statements:
\(f'^{-1} (S) = f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\)
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2: Note
Especially, when \(S \subseteq S_2\), \(f'^{-1} (S) = f^{-1} (S)\), because \(S \cap S_2 = S\) and \(S \setminus S_2 = \emptyset\).
3: Proof
Whole Strategy: Step 1: see that \(f'^{-1} (S) \subseteq f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\); Step 2: see that \(f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2) \subseteq f'^{-1} (S)\); Step 3: conclude the proposition.
Step 1:
For each \(p \in f'^{-1} (S)\), \(f' (p) \in S\), but while \(p \in S_1\) or \(p \in S'_1 \setminus S_1\), when \(p \in S_1\), \(f' (p) = f (p) \in S \cap S_2\), so, \(p \in f^{-1} (S \cap S_2)\), and when \(p \in S'_1 \setminus S_1\), \(f' (p) \in S \setminus S_2\), so, \(p \in f'^{-1} (S \setminus S_2)\), so, \(p \in f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\).
So, \(f'^{-1} (S) \subseteq f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\).
Step 2:
For each \(p \in f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\), while \(p \in f^{-1} (S \cap S_2)\) or \(p \in f'^{-1} (S \setminus S_2)\), when \(p \in f^{-1} (S \cap S_2)\), \(f (p) = f' (p) \in S \cap S_2 \subseteq S\), so, \(p \in f'^{-1} (S)\), and when \(p \in f'^{-1} (S \setminus S_2)\), \(f' (p) \in S \setminus S_2 \subseteq S\), so, \(p \in f'^{-1} (S)\).
So, \(f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2) \subseteq f'^{-1} (S)\).
Step 3:
So, \(f'^{-1} (S) = f^{-1} (S \cap S_2) \cup f'^{-1} (S \setminus S_2)\).
