624: For Map, if There Is Inverse Direction Map Which After Original Map Is Identity, Original Map Is Injective

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description/proof of that for map, if there is inverse direction map which after original map is identity, original map is injective

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Topics

About: set

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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

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Starting Context

• The reader knows a definition of injection.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$f_1$: $S_1\to S_2$, $\in \{\text{ the maps }\}$
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Statements:
$\mathrm{\exists }{f}_2:S_2\to S_1(f_2\circ f_1=id)$
$⟹$
$f_1\in \{\text{ the injections }\}$
//

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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 $f_2\circ f_1=id$.

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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.

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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 $p_1\ne p_2$, 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 $p_1\ne p_2$.

So, $f_1$ is injective.

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References

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