description/proof of that for set endomorphism, if there is surjection onto domain s.t. surjection is endomorphism after surjection, endomorphism is identity
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of %structure kind name% endomorphism.
- The reader knows a definition of surjection.
Target Context
- The reader will have a description and a proof of the proposition that for any set endomorphism, if there is any surjection onto the domain of the endomorphism such that the surjection is the endomorphism after the surjection, the endomorphism is the identity.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S \to S\)
//
Statements:
\(\exists S' \in \{\text{ the sets }\}, \exists f': S' \to S \in \{\text{ the surjections }\} (f' = f \circ f')\)
\(\implies\)
\(f = id: S \to S\)
//
3: Proof
Whole Strategy: Step 1: take any \(s \in S\) and see that \(s = f' (s')\) for an \(s' \in S'\); Step 2: see that \(f (s) = s\).
Step 1:
Let \(s \in S\) be any.
As \(f'\) is surjective, there is an \(s' \in S'\) such that \(s = f' (s')\).
Step 2:
As \(f' = f \circ f'\), \(s = f' (s') = f \circ f' (s') = f (s)\).
As \(s\) is arbitrary, that means that \(f\) is the identity map.
3: Note
For the case that \(f\) is a vectors space endomorphism, there is a proposition that requires less for \(f'\).
