Quotient Mapping is Injection iff Equality

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Theorem

Let $\mathcal R$ be an equivalence relation on $S$.

Then the quotient mapping $q_{\mathcal R}: S \to S / \mathcal R$ is an injection if and only if $\mathcal R$ is the equality relation.


Proof

Let $\eqclass x {\mathcal R}, \eqclass y {\mathcal R} \in S / \mathcal R$


Sufficient Condition

Let $q_{\mathcal R}: S \to S / \mathcal R$ be an injection.


Then:

\(\displaystyle x\) \(\mathcal R\) \(\displaystyle y\)
\(\displaystyle \eqclass x {\mathcal R}\) \(=\) \(\displaystyle \eqclass y {\mathcal R}\) Definition of Equivalence Class
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {q_{\mathcal R} } x\) \(=\) \(\displaystyle \map {q_{\mathcal R} } y\) Definition of Quotient Mapping
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y\) Definition of Injection


That is:

$\mathcal R$ is the equality relation.

$\Box$


Necessary Condition

Let $q_{\mathcal R}: S \to S / \mathcal R$ be a mapping which is specifically not an injection.


Then:

\(\, \displaystyle \exists a, b \in S, a \ne b: \, \) \(\displaystyle \map {q_{\mathcal R} } a\) \(=\) \(\displaystyle \map {q_{\mathcal R} } b\) $q_{\mathcal R}$ is not an injection
\(\displaystyle \leadsto \ \ \) \(\displaystyle \eqclass a {\mathcal R}\) \(=\) \(\displaystyle \eqclass b {\mathcal R}\) Definition of Quotient Mapping
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(\mathcal R\) \(\displaystyle b\) Definition of Equivalence Class

That is:

$a \ne b$

but:

$a \mathrel {\mathcal R} b$

and so $\mathcal R$ is not the equality relation.


From Rule of Transposition it follows that:

if $\mathcal R$ is the equality relation then $q_{\mathcal R}$ is an injection.

$\blacksquare$


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