Well-defined and Equivalence relations

Question

I am wondering why the following is well-defined...

The definition of well-defined is given as; $g:(X/\sim) \to Z$ is well-defined if a mapping $f:X \to Z$ can be found where $f$ has the property $x \sim y \implies f(x)=f(y)$

So, I am asked if the following is well-defined; $E(x) \to e^{2i \pi x}$

I thought No. Because, actually, the mapping didn't really make sense to me. Say $x_1,x_2 \in E(x)$ then, $x_1-x_2=z \in \mathbb{Z}$. Now, $E(x_1)=e^{2i \pi (z+x_2)}=e^{2i \pi z}e^{2i \pi x_2}$ and $E(x_2)=e^{2i \pi x_2}$. Thus $E(x_1) \neq E(x_2)$ unless $x_1 = x_2$ so at least, $x_1 \sim x_2$ does not imply $E(x_1) = E(x_2)$. Which is strange, since both $x_1,x_2$ are related and so in the same equivalence class, but the mapping from the said class does not go to the same element in the image.

And how am I supposed to find "mapping $f:X \to Z$ where $f$ has the property $x \sim y \implies f(x)=f(y)$"? I can't randomly try out every possible mapping a person can think of from $X$ to $Z$ of course, but I mean the definition just says "mapping" which is very very broad. Say, even if I construct a mapping $f:X \to e^{2i \pi x}$ such that $f(x_1) = e^{2i \pi x_1}$, for $f(x_1)=f(x_2)$ to happen, I definitely need $x_1=x_2$ for reals...

But the answer apparently says that they are well-defined, and, by definition, I have no idea why.

Am I misunderstanding the definition of well-defined? I really don't see why this mapping is so at all...

Answer 1

Normally well-defined means that if you have choices to make during the evaluation of some function or quantity, it doesn't matter what choices you make you'll end up with the same result. When dealing with equivalence classes this means you have to get the same function value for whatever representative you choose from the equivalence class.

For what you're doing, I believe this should be interpreted as: If $x\sim y\implies f(x)=f(y)$, you have to show that the implication holds for all pairs of elements in the equivalence class. That way, the function can be interpreted as acting on the entire class, rather than just members of the class.

Your example $E(x)$ works, because if $x\sim y$, then $x-y\in\mathbb{Z}$. So, $x=y+n$. Then

$$E(x)=e^{2\pi ix}=e^{2\pi i(y+n)}=e^{2\pi iy}e^{2\pi in} = e^{2\pi iy}=E(y)$$

This works because $e^{2\pi in}=1$ for any integer $n$.

So we just showed that $x\sim y\implies E(x)=E(y)$. Because $x$ and $y$ were arbitrary members of the equivalence class, this is well-defined.

Answer 2

'Well-defined' is a rather foggy concept, and you'll find it used in lots of different ways throughout mathematics. Typically, one makes a definition, and then one sometimes has to check various things hat show that the definition makes sense. For example, it's not unheard of to use well-defined in the following sense:

We define a function $f\colon \mathbb N \to \mathbb N$ by setting $f(k)$ to be the smallest natural number $s$ such that every positive integer can be expressed as the sum of at most $s$ $k$-th powers.

Indeed, by the Hilbert-Waring Theorem, such an $s$ exists for every $k$. So this map is well-defined.

As another example, we might define a function $f\colon X\to Y$ by $f(x)=$ such and such, and then prove that the function $f$ is 'well-defined' by showing that our value of $f(x)$ actually is contained in $Y$.

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Now, you are using 'well-defined' in a rather specific way. Indeed, this is the most common usage of the term 'well-defined'. Namely, given a set $X$ and an equivalence relation $\sim$ on $X$, we sometimes define a map $f\colon X/\sim\to Y$ by setting

$f([x])=$ some expression depending on $x$

In other words, there is a function $g\colon X\to Y$ and we have set $f([x])=g(x)$.

Of course, this definition doesn't make sense if there exist $x\sim y$ with $g(x)\ne g(y)$. hen $f([x])=g(x)\ne g(y)=f([y])=f([x])$, which is absurd. So, in order to check that this map is 'well-defined', we need to check that $g(x)=g(y)$ whenever $x\sim y$.

Algebraists prefer to express this as a universal property. Given a set $X$ and an equivalence relation $\sim$ on $X$, there is a natural map $p\colon X\to X/\sim$, given by $p(x)=[x]$. $p$ satisfies the property that $p(x)p(y)$ whenever $x\sim y$, and it is said to be universal among all maps with this property: that is, whenever $g\colon X\to Y$ is a map such that $g(x)=g(y)$ whenever $x\sim $, then there is a unique map $f\colon X/\sim\to Y$ such that $g=f\circ p$.