# Let $\sim$ be a relation on $\mathbb R^2$

Let $\sim$ be the relation on $\mathbb R^2$ defined by: $$(x_1, y_1) ∼ (x_2, y_2) \iff x_1^2 + y_1^2=x_2^2+y_2^2$$ Show that $\sim$ is an equivalence relation on $\mathbb R^2$.

Consider the function $f \colon [0, \infty) \to \mathbb R^2/ \sim$ defined by $f(x) = [(x, 0)]$ $\forall x \in [0, \infty)$. Show that $f$ is bijective.

I have shown it is an equivalence relation and that $f$ is injective but I am now stuck on proving it is surjective. I really am not sure how to begin because it seems obvious it is true? Any hints appreciated

• You should show that every partition has a representative of the form $(x,0)$. – Joe Johnson 126 Nov 4 '16 at 2:41

It is easy to see that the given relation is injective.

For surjectivity, you are supposed to show that for every $[(a,b)] \in \mathbb{R}^2/\sim$, we have that $[(a,b)] = f(x)$ for some $x$.

This is very easy to see: let $x = \sqrt{a^2 + b^2}$. Then, note that $(x,0) \sim (a,b)$, since $x^2 + 0^2 = a^2 + b^2$. Hence, $f(x) = [(x,0)] = [(a,b)]$ as $(x,0) \sim (a,b)$.

Hence, with surjectivity and injectivity done, you can say that $f$ is bijective.

Alternately, you can define an inverse function $g : (\mathbb{R}^2/\sim) \to [0,\infty)$, by $g([a,b]) = \sqrt{a^2+b^2}$.

Note that $g \circ f(x) = g([(x,0)]) = \sqrt{x^2 + 0^2} = x$ for all $x \in [0,\infty)$.

Further, $f \circ g([(a,b)]) = f(\sqrt{a^2+b^2}) = [(\sqrt{a^2+b^2},0)] = [(a,b)]$ as $(\sqrt{a^2+b^2},0) \sim (a,b)$.

Hence, $f \circ g$ and $g \circ f$ are both the identity, hence $g$ is a well defined inverse of $f$, hence $f$ is bijective.

Surjectivity is, indeed, the most "obvious" of the three tasks here. You need to show that every equivalence class is achieved by $f(x)$ for some $x$. To do this, it suffices to show that every element of $\Bbb R^2$ is equivalent (in the $\sim$ sense) to $(x,0)$, where $x\in[0,\infty)$. [Also, you should convince yourself that this task really is sufficient...]

This isn't quite trivial: since you already know the function is injective, you know that you can't make the domain of $f$ any smaller and still get a surjective function. So you might ask why $[1,\infty)$ isn't good enough, and even $(0,\infty)$ isn't good enough.