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