Equivalence Relations: Understanding Compatibility

Question

Synopsis

My textbook, near the end of the section on equivalence relations, mentions the problem of "defining functions on a quotient set". Specifically, assume that $R$ is an equivalence relation on $A$ and that $F: A \rightarrow A$. We ask if there exists a corresponding function $\hat{F}: A/R \rightarrow A/R$ such that for all $x \in A$, $$\hat{F} ([x]_{R}) = \ [F(x)]_R.$$ After introducing this notion, he goes on to state that $\hat{F}$ is ill-defined unless $(xRy) \Rightarrow (F(x) R F(y))$. If this requirement is satisfied, then the function $F$ is "compatible" with $R$

Questions

I have a few questions on this topic. Most of them relate to the conceptual understanding of it.

(1) Why is such a question important? What applications will come from such a function $\hat{F}$? Why does he place so much importance on it (he stars it and spends much more time on it than on other sections)? How could I visualize or understand such a function more intuitively?

(2) What does it mean for something to be "well-defined"? I'm guessing it means that the something, or function in this case, is applicable to the definition, but I'm not entirely sure.

(3) What's so special about these compatible functions besides being well-defined?

Answer 1

First consider the following "function" on rational numbers:

$(*)$ Send $p\over q$ to $p+q$.

This is of course nonsense, since for example $${1\over 2}={2\over 4}\quad\mbox{but}\quad 1+2\not=2+4.$$ That is, $(*)$ may look like the definition of a function on $\mathbb{Q}$, but it actually isn't.

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We can describe this issue using equivalence relations and compatibility.

The idea is that our $(*)$ - which turned out to be nonsense - was described in terms of names for rationals, that is specific choices of numerator/denominator. Different names for the same rational led to different outputs, which is why $(*)$ is totally bogus.

What's going on is that $\mathbb{Q}$ is a quotient set. Specifically:

• Let $X=\mathbb{Z}\times\mathbb{Z}_{\not=0}$ be the set of ordered pairs of integers whose second coordinate is nonzero.

• Let $\sim$ be the equivalence relation on $X$ given by $$(a,b)\sim (c,d)\quad\iff\quad ad=bc.$$

Intuitively, we think of $X$ as the set of names for rationals (with $(a,b)$ standing for $a\over b$), and the equivalence relation $\sim$ tells us when two elements of $X$ "mean the same thing."

Now consider the following:

(**) Send $(p,q)$ to $p+q$.

This defines a perfectly good function with domain $X$. It also looks more-or-less like $(*)$, which was nonsense. The point is that $(**)$ is not compatible with the equivalence relation $\sim$. Every function with domain $X$ which is compatible with $\sim$ "turns into" a function with domain $\mathbb{Q}$ (thought of as $X/\sim$), but every function with domain $X$ which isn't compatible with $\sim$ runs into the same problem as $(**)$: while it makes sense on $X$, it doesn't make sense on $\mathbb{Q}$.

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"Well-defined" and "ill-defined" are just slightly sloppy shorthand. For example:

• We should say "$(*)$ does not describe a function.

• Instead, we often say "The function sending $p\over q$ to $p+q$ is not well-defined."

Similarly, we should say "The sentence "Send $p\over q$ to $p^2\over q^2$" defines a function," but instead we often say "The function sending $p\over q$ to $p^2\over q^2$ is well-defined.

Answer 2

Update

Thanks to some help from @J.W. Tanner and @MJD, I realized that the usefulness of the function $\hat{F}$ described above is that it is precisely the notion of something being "well-defined", mainly that the function when applied to two values in the same equivalence class will produce two outputs in the same equivalence class, allowing us to create operations on a set.