Set of equivalence classes equivalent to preimage of image of collapsing map proof

Question

It says in a book I'm reading on topology that if $\mathit{R}$ is an equivalence relation on a space $X$, $p$ is the collapsing map $x \mapsto [x]$ and $A \subseteq X$ then:

$$x \in A, y \mathit{R}x \Rightarrow y\in A$$ is the same as: $$A=p^{-1}(p(A))$$

And that the proof is easy. But, whilst seeing that it's fairly obvious, I'm having trouble coming up with a proof that doesn't get quite wordy. Can anyone help?

Answer 1

By definition, $A \subset p^{-1}[p[A]]$ for any $A$, so the other implication is the interesting one.

Suppose $A$ satisfies the first closure under $R$ condition. Let $z$ be in $p^{-1}[p[A]]$, which means by definition that $p(z) \in p[A]$, or there exists $a \in A$ such that $p(z) = p(a)$. As $p$ is the collapsing map, this just means that $zRa$, i.e. they define the same class. Now apply the condition (with $x = a, y = z$) to conclude that indeed $z \in A$ and we are done.

If $A = p^{-1}[p[A]]$, let $x \in A$ and $y \in X$ be such that $yRx$. Now $p(y) = p(x)$ by the last, and $p(x) \in p[A]$, so $y \in p^{-1}[p[A]] = A$, and so we are done.

The basis idea is that $p(x) = p(y)$ iff $yRx$, and then applying the definitions.

Answer 2

First, note that it does not have anything to do with topology. You can take $X$ to be just a set, $\mathrel{\cal R}$ an equivalence relation on $X$, and $p \colon X \to X\mathop{/}{\cal R}$ the canonical projection.

Then a subset $A \subseteq X$ such that for any $a \in A$, $p(a) \subseteq A$, is called a saturated part of $X$ with respect to $\cal R$.

In other words, a saturated part $A$ is a subset satisfying : if $a \in A$, then $A$ contains all the conjugates of $a$.

And your job is to show that saturated part are exactly those parts $A$ such that $p^{-1}(p(A)) = A$.

You first have to wonder about what is $p^{-1}(B)$ for some $B \subseteq X\mathop{/}{\cal R}$. By definition, it is all of those $x \in X$ whose class modulo $\cal R$ is in $B$. But, by surjectivity of $p$, $B$ is necessarily of the form $p(A)$ for some $A \subseteq X $. So $p^{-1}(B)$ is composed of all those $x \in X$ whose class is in $p(A)$ : it exactly means that $p^{-1}(B)$ is the set of all those $x \in X$ which are in relation with some element of $A$. Sum up this paragraph as :

For any $A \subseteq X$, $p^{-1}(p(A))$ is the subset of $X$ whose elements are all the conjugates of elements of $A$. Formally, $$ p^{-1}(p(A)) = \{x \in X : \exists a \in A , \, x \mathrel{\cal R} a \}.$$

(Remark that $p^{-1}(p(A))$ is actually the smallest saturated subset of $X$ containing $A$ : this is the saturated part generated by $A$.)

The two gray sentences should give you what you need.