# 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 $\phi \circ q_\RR = f$

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

- $\begin {xy} \[email protected] + [email protected] + 1em { S \ar[r]^*{f} \ar[d]_*{q_\RR} & T \\ S / \RR \[email protected]{-->}[ur]_*{\phi} } \end {xy}$

### Mapping from Quotient Set when Defined is Unique

Let the mapping $\phi: S / \RR \to T$ defined as:

- $\phi \circ q_\RR = f$

be well-defined.

Then $\phi$ is unique.

### Condition for Mapping from Quotient Set to be Injection

Let the mapping $\phi: S / \RR \to T$ defined as:

- $\phi \circ q_\RR = f$

be well-defined.

Then:

- $\phi$ is an injection

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

### Condition for Mapping from Quotient Set to be Surjection

Let the mapping $\phi: S / \RR \to T$ defined as:

- $\phi \circ q_\RR = f$

be well-defined.

Then:

- $\phi$ is a surjection

- $f$ is a surjection.

## Motivation

Suppose we are given a mapping $f: S \to T$.

Suppose we have an equivalence $\RR$ on $S$, and we want to define a mapping on the quotient set:

- $\phi: S / \RR \to T$

such that:

- $\map \phi {\eqclass \cdots \RR} = \map f \cdots$

That is, we want every element of a given equivalence class to map to the same element of the codomain of $f$.

The only way this can be done is to set $\map \phi {\eqclass x \RR} = \map f x$.

Now, if $x, y \in S$ are in the same equivalence class class with respect to $\RR$, that is, in order for $\map \phi {\eqclass x \RR}$ to make any sort of sense, we need to make sure that $\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR}$, or (which comes to the same thing) $\map f x = \map f y$.

So $\map \phi {\eqclass x \RR} = \map f x$ defines a mapping $\phi: S / \RR \to T$ if and only if $\forall \tuple {x, y} \in \RR: \map f x = \map f y$.

If this holds, then the mapping $\phi$ is **well-defined**.

The terminology is misleading, as $\phi$ cannot be defined at all if the condition is *not* met.

What this means is: if we want to define a mapping from a quotient set to any other set, then *all* the individual elements of each equivalence class in the domain must map to the *same* element in the codomain.

Therefore, when attempting to construct or analyse such a mapping, it is necessary to check for **well-definedness**.

## Proof

From Condition for Composite Mapping on Left, we have:

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

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

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.

$\blacksquare$

## Also see

## 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 - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Equivalence Relations: $\S 19$ - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.5$