Conditions for Commutative Diagram on Quotient Mappings between Mappings
Jump to navigation
Jump to search
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 $g \circ q_S = q_T \circ f$
- $\forall x, y \in S: \tuple {x, y} \in \RR_S \implies \map {\paren {q_T \circ f} } x = \map {\paren {q_T \circ f} } y$
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 |
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Problem $\text{CC}$: Factoring through Quotients