From ProofWiki
Jump to navigation Jump to search


Well-Defined Mapping

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:

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

Well-Defined Relation

The concept can be generalized to include the general relation $\mathcal R: S \to T$.

Well-Defined Operation

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\mathcal R$ be a congruence for $\circ$.

Let $\circ_\mathcal R$ be the operation induced on $S / \mathcal R$ by $\circ$.

Let $\struct {S / \mathcal R, \circ_\mathcal R}$ be the quotient structure defined by $\mathcal R$, where $\circ_\mathcal R$ is defined as:

$\eqclass x {\mathcal R} \circ_\mathcal R \eqclass y {\mathcal R} = \eqclass {x \circ y} {\mathcal R}$

Then $\circ_\mathcal R$ is well-defined (on $S / \mathcal R$) if and only if:

$x, x' \in \eqclass x {\mathcal R}, y, y' \in \eqclass y {\mathcal R} \implies x \circ y = x' \circ y'$


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

Suppose we have an equivalence $\mathcal R$ on $S$, and we want to define a mapping on the quotient set: $\phi: S / \mathcal R \to T$ such that $\phi \left({\left[\!\left[{...}\right]\!\right]_\mathcal R}\right) = f \left({\ldots}\right)$. That is, we want every member of the equivalence class to map to the same element.

The only way this can be done is to set $\phi \left({\left[\!\left[{x}\right]\!\right]_\mathcal R}\right) = f \left({x}\right)$.

Now, if $x, y \in S$ are in the same equivalence class with respect to $\mathcal R$, that is, in order for $\phi \left({\left[\!\left[{x}\right]\!\right]_\mathcal R}\right)$ to make any sort of sense, we need to make sure that $\phi \left({\left[\!\left[{x}\right]\!\right]_\mathcal R}\right) = \phi \left({\left[\!\left[{y}\right]\!\right]_\mathcal R}\right)$, or (which comes to the same thing) $f \left({x}\right) = f \left({y}\right)$.

So $\phi \left({\left[\!\left[{x}\right]\!\right]_\mathcal R}\right) = f \left({x}\right)$ defines a mapping $\phi: S / \mathcal R \to T$ iff $\forall \left({x, y}\right) \in \mathcal R: f \left({x}\right) = f \left({y}\right)$.

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

The terminology is misleading, as $\phi$ can not 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.

Similarly, when defining an operation on a quotient structure, we need to make sure that the result of performing that operation on two equivalence classes does not depend on the specific instances of the elements in those classes which are chosen to represent them.

Also known as

Some sources use the term consistent for well-defined.