# equivalence relation, quotient space

• January 16th 2009, 05:51 PM
xianghu21
equivalence relation, quotient space
(a) Deﬁne an equivalence relation on the plane $X=\mathbb{R}^2$ as follows:
$x_0 \times y_0$ ~ $x_1 \times y_1$ if $x_0 + y_0^2=x_1 + y_1^2.$
Let $X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set $g(x \times y)=x+y^2.$]

(b) Repeat (a) for the equivalence relation
$x_0 \times y_0$ ~ $x_1 \times y_1$ if $x_0^2+y_0^2=x_1^2+y_1^2.$
[this is Munkres §22.4]
• January 17th 2009, 03:40 AM
HallsofIvy
The quotient space defined by an equivalence relation has, as points, the equivalence classes. Further, a set of such equivalence classes is open if and only if their union is open in the original space.

Here, I notice that, since (0,0) satisfies $x+ y^2= 0+ 0^2= 0$ the equivalence class of (0,0) contains all (x,y) such that $x+ y^2= 0$, a parabola. In fact, given any point $(x_0,y_0)$, its equivalence class consists of all points on the parablola $x+ y^2= x_0+ y_0^2$ Every such parabola has a unique vertex, $(x_0+y_0^2, 0)$. I haven't worked out the details (I'll leave that to you) but I suspect that the function that identifies each such equivalence class with its vertex is a homeomorphism and so the quotient space is homeomorphic to a straight line.

Similarly, for (b), (x, y) is equivalent to a given point $(x_0, y_0)$ if and only if $x^2+ y^2= x_0^2+ y_0^2$, a circle with center (0,0) and radius $\sqrt{x_0^2+ y_0^2}$. Thus every such equivalence class can be identified with the non-negative number $\sqrt{x_0^2+ y_0^2}$ and so the quotient space is homeomorphic to the half open interval $[0, \infty)$.

Your job, now, is to show that those identifications are homeomorphisms.
• January 20th 2009, 01:44 PM
GaloisTheory1
Another way to prove this use:

if f $: X \rightarrow Y$ is a quotient map, then the relation $x$~ $y$ iff $f(x)=f(y)$ is an equivalence relation on $X$, and $Y$ is homeomorphic to $X/$~. I leave the details to you. But either way works.