772: Map Is Bijection iff Preimage of Codomain Point Is 1 Point Subset

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description/proof of that map is bijection iff preimage of codomain point is 1 point subset

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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: Proof

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Starting Context

• The reader knows a definition of bijection.

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Target Context

• The reader will have a description and a proof of the proposition that any map is a bijection iff the preimage of each codomain point is a 1 point subset.

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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:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$f$: $:S_1\to S_2$
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Statements:
$f\in \{\text{ the bijections }\}$
$⟺$
$\mathrm{\forall }s\in S_2(|f^{-1}(s)|=1)$, where $|\bullet |$ is the cardinality of the set
//

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2: Natural Language Description

For any sets, $S_1,S_2$, and any map, $f:S_1\to S_2$, $f$ is a bijection if and only for each $s\in S_2$, $|f^{-1}(s)|=1$.

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3: Proof

Whole Strategy: Step 1: suppose that $f$ is a bijection and see that $|f^{-1}(s)|=1$; Step 2: suppose that $|f^{-1}(s)|=1$ and see that $f$ is a bijection.

Step 1:

Let us suppose that $f$ is a bijection.

Let $s\in S_2$ be any.

$f^{-1}(s)\ne \mathrm{\varnothing }$, because as $f$ is surjective, there is at least 1 point in $S_1$ that (the point) is mapped to $s$, which means that $0<|f^{-1}(s)|$. $|f^{-1}(s)|<2$, because if $2\le |f^{-1}(s)|$, there would be some $s_1,s_2\in S_1$ such that $f(s_1)=f(s_2)=s$, a contradiction against $f$'s being injective. So, $0<|f^{-1}(s)|<2$, which implies that $|f^{-1}(s)|=1$.

Step 2:

Let us suppose that for each $s\in S_2$, $|f^{-1}(s)|=1$.

$f$ is surjective, because there is at least 1 point in $S_1$ that (the point) is mapped to $s$. $f$ is injective, because if there were some $s_1,s_2\in S_1$ such that $f(s_1)=f(s_2)$, $2\le |f^{-1}(f(s_1))|$, a contradiction against $|f^{-1}(s)|=1$ for $s=f(s_1)=f(s_2)$. So, $f$ is a bijection.

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References

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