Quotient Mapping is Injection iff Equality
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Theorem
Let $\RR$ be an equivalence relation on $S$.
Then the quotient mapping $q_\RR: S \to S / \RR$ is an injection if and only if $\RR$ is the equality relation.
Proof
Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$
Sufficient Condition
Let $q_\RR: S \to S / \RR$ be an injection.
Then:
\(\ds x\) | \(\RR\) | \(\ds y\) | ||||||||||||
\(\ds \eqclass x \RR\) | \(=\) | \(\ds \eqclass y \RR\) | Definition of Equivalence Class | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {q_\RR} x\) | \(=\) | \(\ds \map {q_\RR} y\) | Definition of Quotient Mapping | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | Definition of Injection |
That is:
- $\RR$ is the equality relation.
$\Box$
Necessary Condition
Let $q_\RR: S \to S / \RR$ be a mapping which is specifically not an injection.
Then:
\(\ds \exists a, b \in S, a \ne b: \, \) | \(\ds \map {q_\RR} a\) | \(=\) | \(\ds \map {q_\RR} b\) | $q_\RR$ is not an injection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass a \RR\) | \(=\) | \(\ds \eqclass b \RR\) | Definition of Quotient Mapping | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\RR\) | \(\ds b\) | Definition of Equivalence Class |
That is:
- $a \ne b$
but:
- $a \mathrel \RR b$
and so $\RR$ is not the equality relation.
From Rule of Transposition it follows that:
- if $\RR$ is the equality relation then $q_\RR$ is an injection.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations