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Thread: equivalence relation, quotient space

  1. #1
    Nov 2008

    equivalence relation, quotient space

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

    (b) Repeat (a) for the equivalence relation
    $\displaystyle x_0 \times y_0 $ ~ $\displaystyle x_1 \times y_1$ if $\displaystyle x_0^2+y_0^2=x_1^2+y_1^2.$
    [this is Munkres 22.4]
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  2. #2
    MHF Contributor

    Apr 2005
    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 $\displaystyle x+ y^2= 0+ 0^2= 0$ the equivalence class of (0,0) contains all (x,y) such that $\displaystyle x+ y^2= 0$, a parabola. In fact, given any point $\displaystyle (x_0,y_0)$, its equivalence class consists of all points on the parablola $\displaystyle x+ y^2= x_0+ y_0^2$ Every such parabola has a unique vertex, $\displaystyle (x_0+y_0^2, 0)$. 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 $\displaystyle (x_0, y_0)$ if and only if $\displaystyle x^2+ y^2= x_0^2+ y_0^2$, a circle with center (0,0) and radius $\displaystyle \sqrt{x_0^2+ y_0^2}$. Thus every such equivalence class can be identified with the non-negative number $\displaystyle \sqrt{x_0^2+ y_0^2}$ and so the quotient space is homeomorphic to the half open interval $\displaystyle [0, \infty)$.

    Your job, now, is to show that those identifications are homeomorphisms.
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  3. #3
    Dec 2008
    Another way to prove this use:

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