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I have an exercise from my professor;

For the function $f(x)=x^2$, for all $x\in \mathbb{R}$, describe the equivalence relation determined by $f$.

So we are working in the set $\mathbb{R}$, so $x\sim y$ if $f(x)=y, \forall x,y \in \mathbb{R}$? I've worked with some equivalence relations before, but the relation has been given to be. We need $x\sim x, x\sim y \Rightarrow y\sim x,$ and $x\sim y, y\sim z \Rightarrow x\sim z$. This doesnt seem right.....

Or are we looking at a case where $x\sim y$ if $f(x)=f(y)?$ Then the partition generated by $f$ would be $\{0,\{-a,a\}\},a\in \mathbb{R^+}?$

Iceman
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    The first is not an equivalence and the second is, so probably the second is meant (and your partition is correct). :) – Andrew D. Hwang Mar 03 '14 at 01:25
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    The second but $a \in {\mathbb R}_+^*$. In genaral if $f:E \to F$ is a map, the relation $x \mathcal R_f y \Leftrightarrow f(x)=f(y)$ is the equivalence determined by $f$ – Mohamed Mar 03 '14 at 01:33
  • @Mohamed, did you mean $_fxR_fy$? – Iceman Mar 03 '14 at 01:52

1 Answers1

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I think your second version is correct. Said differently, the equivalence classes are exactly the sets $f^{-1}(y)$ for $y\in \text{Im}(f)$, that is, for $y\geq 0$.

MPW
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