# Definition:Well-Defined/Mapping

## Definition

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

Let $\RR$ be an equivalence relation on $S$.

Let $S / \RR$ be the quotient set determined by $\RR$.

Let $\phi: S / \RR \to T$ be a mapping such that:

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

Then $\phi: S / \RR \to T$ is **well-defined** if and only if:

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

## 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**.

## Also known as

Some sources use the term **consistent** for **well-defined**.

Some sources do not hyphenate: **well defined**.

## Examples

### Square Function on Congruence Modulo $6$

Let $x \mathrel {C_6} y$ be the equivalence relation defined on the natural numbers as congruence modulo $6$:

- $x \mathrel {C_6} y \iff x \equiv y \pmod 6$

defined as:

- $\forall x, y \in \N: x \equiv y \pmod 6 \iff \exists k, l, m \in \N, m < 6: 6 k + m = x \text { and } 6 l + m = y$

Let $\eqclass x {C_6}$ denote the equivalence class of $x$ under $C_6$.

Let $\N / {C_6}$ denote the quotient set of $\N$ by $C_6$.

Let us define the mapping $\mathrm {sq}$ on $\N / {C_6}$ as follows:

- $\map {\mathrm {sq} } {\eqclass x {C_6} } = m^2$

where $x = 6 k + m$ for some $k, m \in \N$ such that $m < 6$.

Then $\mathrm {sq}$ is a well-defined mapping.

### Floor of Half Function on Congruence Modulo $6$

Let $x \mathrel {C_6} y$ be the equivalence relation defined on the natural numbers as congruence modulo $6$:

- $x \mathrel {C_6} y \iff x \equiv y \pmod 6$

defined as:

- $\forall x, y \in \N: x \equiv y \pmod 6 \iff \exists k, l, m \in \N, m < 6: 6 k + m = x \text { and } 6 l + m = y$

Let $\eqclass x {C_6}$ denote the equivalence class of $x$ under $C_6$.

Let $\N / {C_6}$ denote the quotient set of $\N$ by $C_6$.

Let us define the mapping $\phi$ on $\N / {C_6}$ as follows:

- $\map \phi {\eqclass x {C_6} } = \eqclass {\floor {x / 2} } {C_6}$

where $\floor {\, \cdot \,}$ denotes the floor function.

Then $\phi$ is *not* a well-defined mapping.

## Also see

- Results about
**well-defined mappings**can be found**here**.

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 6.2$. Mappings of quotient sets - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Equivalence Relations: $\S 19$ - 1996: Winfried Just and Martin Weese:
*Discovering Modern Set Theory. I: The Basics*... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions