Quotient Theorem for Surjections

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Theorem

Let $f: S \to T$ be a surjection.


Then there is one and only one bijection $r: S / \mathcal R_f \to T$ such that:

$r \circ q_{\mathcal R_f} = f$

where:

$\mathcal R_f$ is the equivalence induced by $f$
$r: S / \mathcal R_f \to T$ is the renaming mapping
$q_{\mathcal R_f}: S \to S / \mathcal R_f$ is the quotient mapping induced by $\mathcal R_f$.


This can be illustrated using a commutative diagram as follows:

$\begin {xy} \[email protected] + [email protected] + 1em { S \[email protected]{-->}[rr]^*{f = r \circ q_{\mathcal R_f} } \ar[dd]_*{q_{\mathcal R_f} } && T \\ \\ S / \mathcal R_f \ar[urur]_*{r} } \end {xy}$


Proof

From the definition of Induced Equivalence, the mapping $f: S \to T$ induces an equivalence $\mathcal R_f$ on its domain.

As $f: S \to T$ is a surjection, $T = \Img f$ by definition.

From Renaming Mapping is Bijection, the renaming mapping $r: S / \mathcal R_f \to T$ is a bijection, where $S / \mathcal R_f$ is the quotient set of $S$ by $\mathcal R_f$.

Hence:

$r \circ q_{\mathcal R_f} = f$
$r$ is the only mapping $r: S / \mathcal R_f \to T$ that satisfies this equality.

$\blacksquare$


Also known as

Also known as the factor theorem for surjections.


Also see


Sources