Show that the values of F are the equivalence classes of the equivalence relation

Question

I have the relation $aRb \iff f(a)=f(b)$ where $f: X \to Y$ which I know is an equivalence relation.

For $y \in f(X)$, define $F(y)=f^{-1}(\{y\})$.

How can I show that the values of F are the equivalence class of the first relation?

Answer 1

First of all for $y\in f(X)$ the set $F(y)$ is not empty. So take $a\in F(y)$ and we'll show that $F(y)=[a]$. If $b\in [a]$ then $aRb$ and that means $f(b)=f(a)=y$. Hence $b\in F(y)$. For the other direction suppose $b\in F(y)$. It means $f(b)=y$. In that case $f(b)=f(a)$ and hence $b\in [a]$. So for each $y\in f(X)$ the set $F(y)$ is indeed an equivalence class of the relation you defined.

Answer 2

The question may be re-phrase as follows.

Let $f:X\rightarrow Y$ be a mapping. $f$ induces an equivalence relation on $X$: Define $R=\{(a,b)\in X^{2}\mid f(a)=f(b)\}$. Then $R$ is an equivalence relation on $X$. Let $X/R$ be the set of all equivalent classes. Prove that $X/R=\{f^{-1}[\{y\}]\mid y\in f[X]\}$.

Proof: Let $A\in X/R$. By definition, $A$ is non-empty. Choose $a\in A$ and define $b=f(a)$. We assert that $A=f^{-1}[\{b\}]$. Let $x\in A$, then $xRa\Rightarrow f(x)=f(a)=b$, so $x\in f^{-1}[\{b\}]$. Therefore $A\subseteq f^{-1}[\{b\}]$. Conversely, let $x\in f^{-1}[\{b\}]$, then $f(x)\in\{b\}\Rightarrow f(x)=b=f(a)\Rightarrow xRa$. Therefore $x\in A$. This shows that $f^{-1}[\{b\}]\subseteq A$. It follows that $A=f^{-1}[\{b\}]$.

The above shows that $X/R\subseteq\{f^{-1}[\{y\}]\mid y\in f[X]\}$. Conversely, let $b\in f[X]$ and define $A=f^{-1}[\{b\}]$. We assert that $A$ is an equivalent class. Since $b\in f[X]$, there exists $a\in X$ such that $b=f(a)$. Note that $a\in A$ and hence $A$ is non-empty. Next, we show that for any $x\in X$, $x\in A$ iff $xRa$. Let $x\in X$. If $x\in A$, then $f(x)\in\{b\}\Rightarrow f(x)=b=f(a)\Rightarrow xRa$. Conversely, if $xRa$, then $f(x)=f(a)=b\Rightarrow x\in f^{-1}[\{b\}]=A$. This shows that $A$ is indeed an equivalence class. Therefore $\{f^{-1}[\{y\}]\mid y\in f[X]\}\subseteq X/R$.