description/proof of that composition of map after preimage is identical if map is surjective w.r.t. argument subset
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
- The reader knows a definition of set.
- The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
Target Context
- The reader will have a description and a proof of 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.
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:
\(\forall p \in S_2 (p \in f (S'_1))\)
\(\implies\)
\(f \circ f^{-1} (S_2) = S_2\)
//
When \(f\) is a surjection, the condition is satisfied for any subset of \(S'_2\).
2: Natural Language Description
For any sets, \(S'_1, S'_2\), any map, \(f: S'_1 \to S'_2\), and any subset, \(S_2 \subseteq S'_2\), if \(f\) is surjective with respect to \(S_2\), which means that any point on \(S_2\) is in the range of \(f\), \(f \circ f^{-1} (S_2) = S_2\).
When \(f\) is a surjection, \(f\) is surjective with respect to any subset of \(S'_2\).
3: Proof
Whole Strategy: Step 1: see that \(f \circ f^{-1} (S_2) \subseteq S_2\) in general; Step 2: suppose that \(\forall p \in S_2 (p \in f (S'_1))\), and see that \(S_2 \subseteq f \circ f^{-1} (S_2)\).
Step 1:
\(f \circ f^{-1} (S_2) \subseteq S_2\), by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
So, it is about \(S_2 \subseteq f \circ f^{-1} (S_2)\).
Step 2:
Let us see that \(S_2 \subseteq f \circ f^{-1} (S_2)\).
Let us suppose that \(\forall p \in S_2 (p \in f (S'_1))\).
For any \(p \in S_2\), there is a \(p' \in f^{-1} (p) \subseteq f^{-1} (S_2)\). \(p = f (p') \in f \circ f^{-1} (p) \subseteq f \circ f^{-1} (S_2)\).
