Condition for Mapping from Quotient Set to be Well-Defined

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 $\RR$.

Then:
 * there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$


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


 * $\begin {xy} \xymatrix@L + 2mu@ + 1em {

S \ar[r]^*{f} \ar[d]_*{q_\RR} & T \\ S / \RR \ar@{-->}[ur]_*{\phi} } \end {xy}$

Proof
From Condition for Composite Mapping on Left, we have:


 * $\exists \phi: S / \RR \to T$ such that $\phi$ is a mapping and $\phi \circ q_\RR = f$


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

But by definition of the quotient mapping induced by $\RR$:
 * $\map {q_\RR} x = \map {q_\RR} y \iff \tuple {x, y} \in \RR$

Hence the result.

Also see

 * Definition:Well-Defined Mapping