description/proof of that preimage under codomain-restricted map is preimage under original map
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map.
Target Context
- The reader will have a description and a proof of the proposition that for any map and its any codomain-restriction, the preimage of any subset of the original codomain under the codomain-restricted map is the preimage of the subset under the original map.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S'_2\): \(\in \{\text{ the sets }\}\)
\(f'\): \(: S_1 \to S'_2\)
\(S_2\): \(\subseteq S'_2\) such that \(f' (S_1) \subseteq S_2\)
\(f\): \(: S_1 \to S_2, s \mapsto f' (s)\)
\(S\): \(\subseteq S'_2\)
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Statements:
\(f^{-1} (S) = f'^{-1} (S)\)
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2: Proof
Whole Strategy: Step 1: for each \(s \in f^{-1} (S)\), see that \(s \in f'^{-1} (S)\); Step 2: for each \(s \in f'^{-1} (S)\), see that \(s \in f^{-1} (S)\).
Step 1:
Let \(s \in f^{-1} (S)\) be any.
\(f (s) \in S\).
But \(f' (s) = f (s) \in S\).
So, \(s \in f'^{-1} (S)\).
Step 2:
Let \(s \in f'^{-1} (S)\) be any.
\(f' (s) \in S\).
But \(f (s) = f' (s) \in S\).
So, \(s \in f^{-1} (S)\).
