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

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

Necessary Condition
Suppose:


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

It remains to be shown that $\phi$ is injective.

That is, it remains to be shown that:


 * $\forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$

By definition, $\phi$ is a mapping from $S$ to $S / \RR$.

Thus:


 * $\forall x \in S, \exists \map f x \in S / \RR: \map \phi {\eqclass x \RR} = \map f x$

And also:


 * $\forall y \in S, \exists \map f y \in S / \RR: \map \phi {\eqclass y \RR} = \map f y$

Combining the results:


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

Then:


 * $\forall x, y \in S, \forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$

By the definition of an equivalence class:


 * $\forall x, y \in S, \forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$

Hence:
 * $\forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$

as required.

Sufficient Condition
Suppose $\phi$ is injective. It remains to be shown that:


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

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

Thus, it remains to be shown that:
 * $\forall x, y \in S: \map f x = \map f y \implies \tuple {x, y} \in \RR$

By definition, $\phi$ is injective :


 * $\forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$

By the definition of an equivalence class:


 * $\forall x, y \in S, \forall \eqclass x \RR, \eqclass y \RR \in S / \RR: \map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$

By the definition of a quotient map:


 * $\forall x, y \in S, \forall \map {q_\RR} x, \map {q_\RR} y \in S / \RR: \map \phi {\map {q_\RR} x} = \map \phi {\map {q_\RR} y} \implies \tuple {x, y} \in \RR$

That is:


 * $\forall x, y \in S: \map \phi {\map {q_\RR} x} = \map \phi {\map {q_\RR} y} \implies \tuple {x, y} \in \RR$

By the definition of a composition of mappings:


 * $\forall x, y \in S, \forall \map {q_\RR} x, \map {q_\RR} y \in S / \RR: \map {\paren {\phi \circ q_\RR} } x = \map {\paren {\phi \circ q_\RR} } y \implies \tuple {x, y} \in \RR$

Hence, by the well-definedness of $\phi \circ q_\RR = f$:


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

as required.

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