$f:X→Y$
$x,y ∈ X,xRy$ iff $f(x) = f(y)$
Show that R is an equivalence relation on X.
Also when $X = Y = \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R} $ with $x \mapsto x^2$ for all $x∈R$ find the equivalence classes of R relative to this situation.
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@HagenvonEitzen i wrote f(x) = f(y) implies that f(x)Rf(y) which implies xRy. i know this is wrong:p – user135688 Sep 15 '14 at 11:52
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The statement '$R$ is reflexive' can be restated as '$f(x)=f(x)$ for each $x$'. This is evidently true. Likewise you can do with symmetry and transitivity. Have a look here – drhab Sep 15 '14 at 11:55
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1This is every equivalence relation. – Did Sep 15 '14 at 11:55
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See also: Prove that the relation on $X$ given by $x\sim y$ if $f(x)=f(y)$ is an equivalence relation – Martin Sleziak Dec 06 '21 at 13:05
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Let us start by the easiest condition - reflexivity. Is $xRx$? The answer is yes because clearly $f(x)=f(x)$.
Transitivity: if $xRy$ and $yRz$ then $f(x)=f(y)$ and $f(y)=f(z)$, but clearly $f(x)=f(y)=f(z)$ and thus $f(x)=f(z)$ which implies $xRz$.
Symmetry is easy, I leave this as an exercise.
Now for $x \mapsto x^2$. We denote $x \sim y \iff x^2 = y^2$. Solving this equation yields $x = \pm y$. Thus $x \sim y \iff x = \pm y$. There are only two such $x$'s and they are $y$ and $-y$ so each class is $[y] = \{ y , - y \}$ and $[0]=\{ 0 \}$.
LinAlgMan
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1$0$ is its own equivalence class, since it is the unique solution of $x^2=0$. – LinAlgMan Sep 17 '14 at 11:35