How can I prove that these two topologies are the same?

In this case, I'm only confused, sorry for ask, but I'm learning the quotient topology ._. I know that the problem it's trivial . Sorry again for ask.

Let $S^2$ and define an equivalence relation , $(x,y,z)$ related with $(-x,-y,-z)$ and define here the quotient topology. Prove that this is homeomorphic to $$R^3 - \left\{ {\left( {0,0,0} \right)} \right\}$$ with the quotient topology given by the relation $$\left( {x,y,z} \right) \sim \left( {tx,ty,tz} \right) .$$ and t not zero EDITED: thanks

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Perhaps you are missing that $t \neq 0$ in the second topology. – Srivatsan Sep 12 '11 at 2:03
There's no reason to apologize for asking a question, nor is there any reason to downplay your questions as "trivial". You've already climbed rather high to be able to view topology at all. – Austin Mohr Sep 12 '11 at 2:49
There's no such thing as a stupid question. Anyone that thinks there is has no business teaching others.That's my take on it.So never be embarrassed to ask one! – Mathemagician1234 Sep 12 '11 at 3:45

First look at it intuitively. The relation $$(x,y,z) \sim_1 (tx,ty,tz)$$ on $\mathbb{R}^3\setminus \{(0,0,0)\}$ collapses each line through the origin to a single point. Each of those lines through the origin intersects the sphere $S^2$ in exactly two points; if one of them is $(x,y,z)$, the other is $(-x,-y,-z)$, so the relation $$(x,y,z) \sim_2 (-x,-y,-z)$$ on $S^2$ collapses them to a single point as well. This point corresponds to the collapse of the line in the first quotient.
You can use this idea to find an actual homeomorphism. Let $q_1:\mathbb{R}^3\setminus \{(0,0,0)\}\to X$ be the first quotient map and $q_2:S^2\to Y$ be the second, where $X = \mathbb{R}^3\setminus \{(0,0,0)\}/\sim_1$ and $Y=S^2/\sim_2$. Let $p$ be any point of $Y$; then there is a point $(x,y,z)\in S^2$ such that $$p = q_2((x,y,z))=q_2((-x,-y,-z)).$$ Clearly $(x,y,z) \ne (0,0,0)$, so $q_1((x,y,z)) \in X$.
Now let $h(p) = q_1((x,y,z))$. Note that $q_1((-x,-y,-z))=q_1((x,y,z))$ (why?), so it doesn’t matter which of the two $q_2$-preimages of $p$ I use. I claim that $h$ is a homeomorphism from $Y$ onto $X$, but I’ll leave you to try to work out the details.
@Daniel: The map $h$ that I described is an explicit homeomorphism between them: for each point of $Y$ it gives you the corresponding point of $X$. – Brian M. Scott Sep 13 '11 at 3:34