description/proof of that for surjection, preimages of subsets are same iff subsets are same
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any surjection, the preimages of any subsets of the codomain are same iff the subsets are same.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(S_1 \to S_2\), \(\in \{\text{ the surjections }\}\)
\(S_{2, 1}\): \(\subseteq S_2\)
\(S_{2, 2}\): \(\subseteq S_2\)
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Statements:
\(f^{-1} (S_{2, 1}) = f^{-1} (S_{2, 2})\)
\(\iff\)
\(S_{2, 1} = S_{2, 2}\)
//
2: Natural Language Description
For any sets, \(S_1, S_2\), any surjection, \(f: S_1 \to S_2\), and any subsets, \(S_{2, 1}, S_{2, 2} \subseteq S_2\), \(f^{-1} (S_{2, 1}) = f^{-1} (S_{2, 2})\) if and only if \(S_{2, 1} = S_{2, 2}\).
3: Proof
Whole Strategy: Step 1: suppose that \(f^{-1} (s_{2, 1}) = f^{-1} (s_{2, 2})\), and see that \(S_{2, 1} = S_{2, 2}\); Step 2: suppose that \(S_{2, 1} = S_{2, 2}\), see that \(f^{-1} (s_{2, 1}) = f^{-1} (s_{2, 2})\).
Step 1:
Let us suppose that \(f^{-1} (S_{2, 1}) = f^{-1} (S_{2, 2})\).
\(f \circ f^{-1} (S_{2, 1}) = f \circ f^{-1} (S_{2, 2})\). But \(f \circ f^{-1} (S_{2, 1}) = S_{2, 1}\) and \(f \circ f^{-1} (S_{2, 2}) = S_{2, 2}\), by the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset. So, \(S_{2, 1} = S_{2, 2}\).
Step 2:
Let us suppose that \(S_{2, 1} = S_{2, 2}\).
\(f^{-1} (S_{2, 1}) = f^{-1} (S_{2, 2})\) is immediate.
