description/proof of that for \(2\) sets, maps from 1st set into 2nd set are same iff preimages of each subset of codomain are same
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 \(2\) sets, any maps from the 1st set into the 2nd set are same if and only if the preimages of each subset of the codomain 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(f'\): \(: S_1 \to S_2\)
//
Statements:
\(f = f'\)
\(\iff\)
\(\forall S \subseteq S_2 (f^{-1} (S) = f'^{-1} (S))\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f = f'\); Step 2: see that \(\forall S \subseteq S_2 (f^{-1} (S) = f'^{-1} (S))\); Step 3: suppose that \(\forall S \subseteq S_2 (f^{-1} (S) = f'^{-1} (S))\); Step 4: suppose that \(f \neq f'\), and find a contradiction.
Step 1:
Let us suppose that \(f = f'\).
Step 2:
\(\forall S \subseteq S_2 (f^{-1} (S) = f'^{-1} (S))\) is obvious (otherwise, the concept of 'preimage' would be ill-defined): to be more specific, \(f^{-1} (S) := \{s_1 \in S_1 \vert f (s_1) \in S\} = \{s_1 \in S_1 \vert f' (s_1) \in S\}\), which is \(f'^{-1} (S)\), by definition: if \(\{s_1 \in S_1 \vert f (s_1) \in S\}\) is doubted to determine a definite set, it will be a denial of the whole set theory.
Step 3:
Let us suppose that \(\forall S \subseteq S_2 (f^{-1} (S) = f'^{-1} (S))\).
Step 4:
Let us suppose that \(f \neq f'\).
There would be an \(s_1 \in S_1\) such that \(f (s_1) \neq f' (s_1)\).
Let \(S = \{f (s_1)\} \subseteq S_2\).
\(s_1 \in f^{-1} (S)\), because \(f (s_1) \in S\).
But \(s_1 \notin f'^{-1} (S)\), because if \(s_1 \in f'^{-1} (S)\), \(f' (s_1) \in S\), which would mean that \(f' (s_1) = f (s_1)\), a contradiction.
That would mean that \(f^{-1} (S) \neq f'^{-1} (S)\), a contradiction.
So, \(f = f'\).
