Condition for Mapping from Quotient Set to be Injection

Theorem
Let $S$ and $T$ be sets.

Let $\RR$ be an equivalence relation on $S$.

Let $f: S \to T$ be a mapping from $S$ to $T$.

Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.

Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$. Let the mapping $\phi: S / \RR \to T$ defined as:
 * $\phi \circ q_\RR = f$

be well-defined.

Then:
 * $\phi$ is an injection


 * $\forall x, y \in S: \tuple {x, y} \in \RR \iff \map f x = \map f y$
 * $\forall x, y \in S: \tuple {x, y} \in \RR \iff \map f x = \map f y$

Proof
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined :
 * $\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$

By definition of injection, $\phi$ is injective :


 * $\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$

that is:
 * $\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$

By definition of $\phi$:
 * $\map \phi {\eqclass x \RR} = \map f x$

and so:
 * $\map f x = \map f y \implies \tuple {x, y} \in \RR$

Hence the result.

Also see

 * Definition:Well-Defined Mapping


 * Condition for Mapping from Quotient Set to be Well-Defined
 * Mapping from Quotient Set when Defined is Unique
 * Condition for Mapping from Quotient Set to be Surjection