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I have a homework question: Consider the maps: $$f : \mathbb{R}^2 \to \mathbb{R},\; (x,y) \mapsto x^2 +y^2$$ $$f : \mathbb{R} \to \mathbb{R},\; x \mapsto x^3$$ $$f : \mathbb{C} \to \mathbb{C},\; z \mapsto z^3$$ Determine the equivalence relation on the respective domains determined by $f$. Namely, explicitly describe equivalence classes from the map $f$.

I have no idea what it is asking, I know that equivalence relations can be used to divvy up a set, and I know that they are reflexive, symmetric, and transitive. But I do not know how to approach this.

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  • $\begingroup$ For the second one the set of equivalence classes be $\{[[0]]_f,[[1]]_f\}$ where $[0],[1]∈Z_2$.Since a cube of a odd number is always odd and a cube of a even number is always even $\endgroup$ – user60887 Sep 18 '13 at 5:30
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HINT: If you have a function $f:X\to Y$, it defines an equivalence relation $\sim$ on $X$ in the following way: for $x_0,x_1\in X$, $x_0\sim x_1$ if and only if $f(x_0)=f(x_1)$. I’ll leave to you the easy verification that this $\sim$ really is an equivalence relation on $X$. The equivalence class of an $x\in X$ is the set of all $y\in X$ such that $f(x)=f(y)$, so it’s just $f^{-1}[\{f(x)\}]$.

In your first problem, for instance, you have $f:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x^2+y^2$; geometrically speaking, that sends a point in the plane to the square of its distance from the origin, so all points that are at the same distance from the origin get sent to the same real number. Thus, the equivalence classes are just circles centred on the origin.

I’ll leave the rest for you now.

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