# Definition talk:Well-Defined

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If an "unambiguous" relation is different from a "well-defined" relation, why the need to disambiguate? --prime mover (talk) 21:19, 6 February 2013 (UTC)

- I've seen "well defined" to mean "unambiguous", like in Definition:Inversion Mapping. Or is that using well-defined like it is on this page? --GFauxPas (talk) 21:27, 6 February 2013 (UTC)

- Strictly speaking, the meaning as on Definition:Inversion Mapping is the same as for quotient relations - but with the admittedly trivial case of the diagonal equivalence relation. Perhaps we could say a "multifunction $f$ is
**well-defined (for $\mathcal R$)**iff $(x,y) \in \mathcal R$ implies $f(x) = f(y)$" where $\mathcal R$ is an equivalence. --Lord_Farin (talk) 22:07, 6 February 2013 (UTC)

- Strictly speaking, the meaning as on Definition:Inversion Mapping is the same as for quotient relations - but with the admittedly trivial case of the diagonal equivalence relation. Perhaps we could say a "multifunction $f$ is

- Okay, so maybe not "disambiguate", but add an "also defined as" to indicate that it means the same as "unambiguous". But you then have to go and define what "unambiguous" means - no such definition exists on this site, and I for one have no intuitive idea as to what it
*might*mean. --prime mover (talk) 22:20, 6 February 2013 (UTC)

- Okay, so maybe not "disambiguate", but add an "also defined as" to indicate that it means the same as "unambiguous". But you then have to go and define what "unambiguous" means - no such definition exists on this site, and I for one have no intuitive idea as to what it

## General Concept

Would it be fair to say that the essential concept here is that the notation:

- Let $a$ be the object such that $P(a)$
- Lemma: $a$ is well-defined.
- Proof: $\cdots \vdash (\exists x: P(x)) \land (\forall y: (P(y) \implies y = x))$

is just another way of saying:

- Lemma: There exists a unique object $x$ such that $P(x)$
- Proof: $\cdots \vdash (\exists x: P(x)) \land (\forall y: (P(y) \implies y = x))$
- Let $a$ be the object such that $P(a)$. (where we presumably do something with uniqueness to allow $a$ to break free from the existential instantiation).

-- Dfeuer (talk) 19:18, 12 March 2013 (UTC)

- Not really.

- It is a specific statement about mappings and operations on equivalence classes.

- What "well-defined" means is: for all elements of a given set, performing operation $f$ on those elements always results in the same result. "$f$ is well-defined on $S$" means: $\exists y: \forall x \in S: f \left({x}\right) = y$.

- In the context within which it is used, $S$ is usually an equivalence class; in its rawest form, that equivalence class is that induced by the mapping. --prime mover (talk) 19:47, 12 March 2013 (UTC)

- More adequate would be, in the usual case of an equivalence relation $\sim$ with quotient map $q: S \to S / \sim$, to say that a proof of well-definedness is a proof that a map $f: S \to T$ factors through $q$, i.e. there is $\tilde f: S / \sim \to T$ such that $f = \tilde f \circ q$. It is usual to define $\tilde f$ in terms of $f$ because of conceptual simplicity. — Lord_Farin (talk) 21:53, 12 March 2013 (UTC)