Conditions for Commutative Diagram on Quotient Mappings between Mappings

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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:

$q_T \circ f = g \circ q_S$

if and only if:

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


$\begin {xy} \xymatrix@L + 2mu@ + 1em { S \ar[r]^*{f} \ar[d]_*{q_S} & T \ar[d]^*{q_T} \\ S / \RR_S \ar@{-->}[r]_*{g} & T / \RR_T } \end {xy}$


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 $g \circ q_S = q_T \circ f$

if and only if:

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


Hence:

\(\ds \map {\paren {q_T \circ f} } x\) \(=\) \(\ds \map {\paren {q_T \circ f} } y\)
\(\ds \leadstoandfrom \ \ \) \(\ds \map {q_t} {\map f x}\) \(=\) \(\ds \map {q_t} {\map f y}\) Definition of Composition of Mappings
\(\ds \leadstoandfrom \ \ \) \(\ds \map f x\) \(\RR_T\) \(\ds \map f y\) Definition of Quotient Mapping

$\blacksquare$


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