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I don't really understand the notion of well defined function. For exemple, I have to prove that $$\Phi: G/G_x\to X$$ define by $[g]=g\cdot x$ is well defined. We have that $G_x=\{g\in G\mid g\cdot x=x\}$. In me course, it's written that $\Phi$ is well define if $$[x]=[y]\implies \Phi([x])=\Phi([y]).$$ I don't really understand this. Indeed, if $a=b$ it's obvious that $\Phi(a)=\Phi(b)$ no ? How can't it be ? I would say that $\Phi$ is well define if $\Phi(x)$ has a sense. For example, if $f(x)=\frac{1}{x}$, then $f$ is not well define on $\mathbb R$, since $f(0)$ is not defined. But I don't understand what they want to say here. Is there a similitude between my intuition and the fact that $a=b\implies \Phi(a)=\Phi(b)$ ?

Thank you.

Rick
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  • http://math.stackexchange.com/questions/606917/well-defined-function-what-does-it-mean – user236182 Oct 07 '15 at 14:27
  • I saw in every document that $f:A\to B$ is well define if for all $a\in A$ there is a unique $b\in B$ such that $b=f(a)$. How can $f:x\mapsto \frac{1}{x}$ is well define from $\mathbb R\to\mathbb R$ since $f(0)\not\in\mathbb R$. Therefore, there is no $a\in \mathbb R$ such that $a=f(0)$. – Rick Oct 07 '15 at 14:36
  • An example of a not well-defined function (so technically not a function) would be any multifunction. – user236182 Oct 07 '15 at 14:40

2 Answers2

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It is indeed wrong to say that a function $f$ on equivalence classes is well defined if $[x]=[y]\implies f([x])=f([y])$.

This statement only makes sense if it is allready known that $f$ is a (well defined) function. Secondly it is true for any function $f$ on the equivalence classes, so it offers no information.

It should be something like this:

If $[x]=[y]\implies g(x)=g(y)$ then the function prescribed by $[x]\mapsto g(x)$ is well defined.

This because the RHS will not be affected if another representative is chosen.

Denoting this function with $f$ we have $f([x]):=g(x)$.

Actually it has been proved that $g=f\circ[]$ where $[]$ denotes the natural map $x\mapsto [x]$.

Or in words that function $g$ can be factored over $[]$.

So proving that this (I would say) "predefined function" $f$ is indeed a well defined function is actually showing that indeed $[x]=[y]\implies g(x)=g(y)$

drhab
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  • Thanks for your answer. How would you adapt it to my problem ? What would be $g$ ? It looks to be $\pi: x\mapsto [x]$, but we have by definition that $[x]=[y]\implies \pi(x)=\pi(y)$. – Rick Oct 07 '15 at 15:37
  • If I understand well then in your case it must be checked wether $gG_x=[g]=[h]=hG_x$ implies that $g.x=h.x$. This is indeed the case because it implies that $h^{-1}g\in G_x$ and from this it can be deduced that $g.x=h.x$. After this check you can say that function $[g]\mapsto g.x$ is a well defined function on $G/G_x$. – drhab Oct 07 '15 at 15:48
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All it says here is that the function is defined on an equivalence class and if you want the function to be defined such that it makes sense i.e given an input you want an output. If [x]=[y] then x may not be the same as y. If your function differentiates between x and y and gives different for x and y then it makes no sense to say that it is defined properly on the class. So what you want is that given an equivalence class, the function should have the same value on every element of it and thus we may talk about the function having a value on the class.

(Technically, the equivalence classes are a different object in itself and the function may not be defined on each element of the set but at the equivalence class only. But that presents no problem. The problem occurs when you have a function on the elements of set and you have some equivalence relation between the elements and you try to combine then up and define a function on the equivalence classes as we often do when defining a map on quotient spaces using a map on the original space which was quotiented.)

(For well definition in general, do have a look at the link suggested in the comment.)

Neeraj Kumar
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