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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:

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


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 $\phi$ 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.

Also known as

Some sources use the term consistent for well-defined.

Also see

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