January 19th, 2018, 03:56 AM  #1 
Newbie Joined: Dec 2017 From: vienna Posts: 8 Thanks: 1  Quotient topology
let X= R^2{(0,0)} with the equivalence relation : (x,y)R(z,w) iff y=w=0 and x/z positive real number or if y=w not equal to 0 . 1can you describe the set of all equivalence classes of X? 2give me a basis of of the quotient topology ? 3can you give a homeomorphism for which this quotient space is homemorphic to ? I have a problem in imagining the topology of the quotient space and how to determine the set of all equivalence classes ,can u provide me with some advices ? Thank u 
January 19th, 2018, 05:38 AM  #2 
Senior Member Joined: Oct 2009 Posts: 227 Thanks: 80 
Start by drawing a picture of $\mathbb{R}^2$ and drawing some equivalence classes in a color.

January 20th, 2018, 08:36 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 273 Thanks: 139 Math Focus: Dynamical systems, analytic function theory, numerics 
1. I think should be pretty straightforward for the level of course you seem to be working in. 2/3. This is where i suspect the trouble lies. I doubt its helpful to give you a full solution but if you are still stuck you may find the following theorem and discussion helpful. First, recall that a map between topological spaces, $g: X \to A$ is a quotient map if $g$ is surjective and any subset, $U \subset A$ is open in $A$ if and only if $g^{1}(U)$ is open in $X$. Note: This definition appears at first glance to simply be the definition of continuity but look closer and you will see that it is quite a bit stronger. Theorem: Suppose $X,A$ are topological spaces, and $A$ is Hausdorff. Let $g: X \to A$ be a continuous surjection and define \[Y = \{g^{1}(a) \ \colon a \in A \} \] endowed with the quotient topology. Then $Y$ is also Hausdorff and $g$ induces a continuous bijection, $\tilde{g}: Y \to A$, and $\tilde{g}$ is a homeomorphism if and only if $g$ is a quotient map. Now, the basic strategy for applying this theorem is the following: (1) Start with an abstract topological space which you want to understand. The goal is to place this in the role of $Y$ in the previous theorem. (2) To do this, you look for an $X$ and a $q$ so that $q: X \to Y$ is precisely the map induced by your identification. Verify that $q$ is a quotient map so that $Y$ has the quotient topology. (3) Determine the topology of $Y$ (this is usually done by experimenting with the equivalence classes and making a good guess) and let $A$ be a typical representative. For example, at a glance my guess for your problem would be that the quotient you have is homeomorphic to the smash product of two torii (i.e. $\mathbb{T}$ ^ $\mathbb{T}$) so take this as an initial guess for $A$. (4) Try to find $g: X \to A$ which is a quotient map satisfying \[q \left( g^{1}(a) \right) = \tilde{g}^{1}(a) \] or equivalently, $g = \tilde{g} \circ q$. Finding this $g$ and verifiying this quotient map is usually straightforward once you have worked out (1)(3). Once found, the theorem above does the work of proving that $A$ is homeomorphic to $Y = X/R$. As a final note, specifically answering 2. of your question is easy to do in the general case. Consider the previous theorem (whose proof is not that difficult and should be sketched at the very least) and look at $g(B)$ where $B$ is a topological basis for $X$. Can you prove that this is a basis for $A$? 

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