(a) Deﬁne an equivalence relation on the plane as follows:

~ if

Let be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set ]

(b) Repeat (a) for the equivalence relation

~ if

[this is Munkres §22.4]

Printable View

- Jan 16th 2009, 05:51 PMxianghu21equivalence relation, quotient space
(a) Deﬁne an equivalence relation on the plane as follows:

~ if

Let be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set ]

(b) Repeat (a) for the equivalence relation

~ if

[this is Munkres §22.4] - Jan 17th 2009, 03:40 AMHallsofIvy
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 the equivalence class of (0,0) contains all (x,y) such that , a parabola. In fact, given any point , its equivalence class consists of all points on the parablola Every such parabola has a unique vertex, . 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 if and only if , a circle with center (0,0) and radius . Thus every such equivalence class can be identified with the**non**-negative number and so the quotient space is homeomorphic to the half open interval .

Your job, now, is to show that those identifications**are**homeomorphisms. - Jan 20th 2009, 01:44 PMGaloisTheory1
Another way to prove this use:

if f is a quotient map, then the relation ~ iff is an equivalence relation on , and is homeomorphic to ~. I leave the details to you. But either way works.