Condition for Mapping from Quotient Set to be Well-Defined

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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 $\phi \circ q_\RR = f$

if and only if:

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


$\begin {xy} \xymatrix@L + 2mu@ + 1em {
S \ar[r]^*{f}
  \ar[d]_*{q_\RR}

& T \\ S / \RR \ar@{-->}[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

That is,

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


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

if and only if:

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

if and only if:

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

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

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