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Let $X$ and $Y$ be two any sets and let $f:X\rightarrow Y$ be a function sucht that $x_{1}\equiv x_{2}\Leftrightarrow f\left( x_{1}\right) =f\left( x_{2}\right)$.

My question is: How can I find equivalence classes of this? I think, the classes are the set of ''önimge'' (''önimge'' is in Turkish, I don't remember English of 'önimge', sorry. But, if $f^{\sim 1}\left( B\right)$ is ''önimge'' than $f^{\sim 1}\left( B\right) =${$ x\in X:f\left( x\right)\in B\subseteq Y$} ).

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    The word you are looking for is [pre-image](http://mathworld.wolfram.com/Pre-Image.html). – 2017-01-09
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    @dxiv Yes, thanks. It is pre-image (önimge). – 2017-01-09

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For a fixed $x_2$ its equivalence class is the set of $x \in X$ such that $f(x) = f(x_2)$ i.e. $f^{-1}(\{f(x_2)\})$ where $f^{-1}(A)$ is the pre-image of set $A\,$ (not to be confused with the inverse function which may or may not exist). Therefore, the set of all equivalence classes is: $\big\{ f^{-1}(\{y\}) \;\;\big|\;\; y \in f(X)\big\}\,$.

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    Thanks but I didn't understand your last sentence. So, what is the set of {$y$} mean? Also, what is the $\left\{ f^{-1}\left(\{ y\right\}) :y\in f\left( X\right) \right\}$ mean? – 2017-01-10
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    $y \in f(X)$ is any element in the range of $f$ which is $f(X) \subseteq Y$ i.e. $y \in f(X)$ $\iff$ $\exists x \in X$ such that $f(x)=y\,$. $f^{-1}(\{y\})$ is the pre-image of the set $\{y\} \subseteq Y$ which contains $y$ as its only element. By the definition of the pre-image $x_1,x_2 \in f^{-1}(\{y\})$ $\iff$ $f(x_1), f(x_2) \in \{y\}$ $\iff$ $f(x_1) = f(x_2) = y$ and by definition of the given equivalence this is the same as $x_1 \equiv x_2$. – 2017-01-10
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If $x\in X$ then, $$[x]=\{y\in X:y\equiv x\}=\{y\in X:f(y)=f(x)\}=f^{-1}(\{z\}), \text{ where }z=f(x).$$