description/proof of that for map, if there is inverse direction map which after original map is identity, original map is injective
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: Note
- 4: Proof
Starting Context
- The reader knows a definition of injection.
Target Context
- The reader will have a description and a proof of the proposition that for any map, if there is any inverse direction map which after the original map is the identity, the original map is injective.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f_1\): \(S_1 \to S_2\), \(\in \{\text{ the maps }\}\)
//
Statements:
\(\exists f_2: S_2 \to S_1 (f_2 \circ f_1 = id)\)
\(\implies\)
\(f_1 \in \{\text{ the injections }\}\)
//
2: Natural Language Description
For any sets, \(S_1, S_2\), any map, \(f_1: S_1 \to S_2\), is injective if there is a map, \(f_2: S_2 \to S_1\), such that \(f_2 \circ f_1 = id\).
3: Note
A typical way of confirming the injectiveness is to check that any 2 distinct points are mapped to distinct points, but also the way of this proposition is useful.
\(f_1\) is not necessarily surjective. As a counterexample, let \(S_1 = \{1\}\), \(S_2 = \{1, 2\}\), and \(f_1: 1 \mapsto 1\). There is \(f_2: 1 \mapsto 1, 2 \mapsto 1\) such that \(f_2 \circ f_1 = id\), but \(f_1\) is not surjective.
4: Proof
Let us suppose that there is any such \(f_2\).
Suppose that \(f_1\) was not injective.
There would be some 2 points, \(p_1, p_2 \in S_1\), such that \(p_1 \neq p_2\), and \(f_1 (p_1) = f_1 (p_2)\). Then, \(f_2 \circ f_1 (p_1) = f_2 \circ f_1 (p_2)\), because \(f_2\) could not map the same point, \(f_1 (p_1) = f_1 (p_2)\), to different points. But \(p_1 = id (p_1) = f_2 \circ f_1 (p_1) = f_2 \circ f_1 (p_2) = id (p_2) = p_2\), a contradiction against that \(p_1 \neq p_2\).
So, \(f_1\) is injective.
