description/proof of that composition of map after preimage is identical iff argument set is subset of map range
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- 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, the composition of the map after any preimage is identical if and only if the argument set is a subset of the map range.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\({S_2}^`\): \(\subseteq S_2\)
//
Statements:
\(f \circ f^{-1} ({S_2}^`) = {S_2}^`\)
\(\iff\)
\({S_2}^` \subseteq f (S_1)\)
//
2: Note
It is important not to carelessly conclude that \(f \circ f^{-1} ({S_2}^`) = {S_2}^`\) without checking the condition.
3: Proof
Whole Strategy: Step 1: suppose that \(f \circ f^{-1} ({S_2}^`) = {S_2}^`\); Step 2: suppose that it was not that \({S_2}^` \subseteq f (S_1)\), and find a contradiction; Step 3: suppose that \({S_2}^` \subseteq f (S_1)\); Step 4: see that for each \(p \in f \circ f^{-1} ({S_2}^`)\), \(p \in {S_2}^`\), and for each \(p \in {S_2}^`\), \(p \in f \circ f^{-1} ({S_2}^`)\).
Step 1:
Let us suppose that \(f \circ f^{-1} ({S_2}^`) = {S_2}^`\).
Step 2:
Let us suppose that it was not that \({S_2}^` \subseteq f (S_1)\).
There would be a \(p \in {S_2}^`\) such that \(p \notin f (S_1)\).
\(p \notin f \circ f^{-1} ({S_2}^`) = {S_2}^`\), a contradiction.
So, \({S_2}^` \subseteq f (S_1)\).
Step 3:
Let us suppose that \({S_2}^` \subseteq f (S_1)\).
Step 4:
For each \(p \in f \circ f^{-1} ({S_2}^`)\), \(p = f (p')\) for a \(p' \in f^{-1} ({S_2}^`)\), which implies that \(f (p') \in {S_2}^`\), so, \(p = f (p') \in {S_2}^`\).
So, \(f \circ f^{-1} ({S_2}^`) \subseteq {S_2}^`\).
For each \(p \in {S_2}^`\), as \({S_2}^` \subseteq f (S_1)\), \(p = f (p')\) for a \(p' \in S_1\), so, \(f (p') = p \in {S_2}^`\), so, \(p' \in f^{-1} ({S_2}^`)\), so, \(p = f (p') \in f \circ f^{-1} ({S_2}^`)\).
So, \({S_2}^` \subseteq f \circ f^{-1} ({S_2}^`)\).
So, \(f \circ f^{-1} ({S_2}^`) = {S_2}^`\).
