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Math Help - equivalence relation, quotient space

  1. #1
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    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]
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  2. #2
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    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.
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  3. #3
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    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.
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