Conditions for Commutative Diagram on Quotient Mappings between Mappings

Theorem
Let $A$ and $B$ be sets.

Let $\RR_S$ and $\RR_T$ be equivalence relations on $S$ and $T$ respectively.

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

Let $S / \RR_S$ and $T / \RR_T$ be the quotient sets of $S$ and $T$ induced by $\RR_S$ and $\RR_T$ respectively.

Let $q_S: S \to S / \RR_S$ and $q_T: T \to T / \RR_T$ be the quotient mappings induced by $\RR_S$ and $\RR_T$ respectively.

Then a mapping $g: S / \RR_S \to T / \RR_T$ exists such that:


 * $q_T \circ f = g \circ q_S$




 * $\forall x, y \in S: x \mathrel {\RR_S} y \implies \map f x \mathrel {\RR_T} \map f y$


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

S \ar[r]^*{f} \ar[d]_*{q_S} & T \ar[d]^*{q_T} \\ S / \RR_S \ar@{-->}[r]_*{g} & T / \RR_T } \end {xy}$

Proof
Consider the commutative diagram:


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

S \ar[rr]^*{f} \ar[dd]_*{q_S} \ar[ddrr]^*{q_T \circ f} & & T \ar[dd]^*{q_T} \\ & & \\ S / \RR_S \ar@{-->}[rr]_*{g} & & T / \RR_T } \end {xy}$

We consider the mapping $q_T \circ f: S \to T / \RR_T$.

From Condition for Mapping from Quotient Set to be Well-Defined:


 * there exists a mapping $g: S / \RR_S \to T / \RR_T$ such that $g \circ q_S = q_T \circ f$


 * $\forall x, y \in S: \tuple {x, y} \in \RR_S \implies \map {\paren {q_T \circ f} } x = \map {\paren {q_T \circ f} } y$
 * $\forall x, y \in S: \tuple {x, y} \in \RR_S \implies \map {\paren {q_T \circ f} } x = \map {\paren {q_T \circ f} } y$

Hence: