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Let $X = \{(x, y) \in R_2 \mid 1 \leq \left||(x, y)\right|| ≤ 2\}$. We define an equivalence relation on $X$ as follows: $(x, y) \sim (x', y')$ if and only if $(x, y) = (x', y')$ or $\left||(x, y )\right|| - \left||(x', y')\right|| = \pm 1$ and $(x, y) = λ (x ', y')$, $λ> 0$. Prove that the quotient space is homeomorphic to the torus.

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you could try to define a homeomorphism between $X/\sim$ and $T^2$. for this you will need to recall how you define $T^2$. – tom b. Mar 13 '13 at 18:51
sorry? Sorry so much but I am new in topology and I have been reading book about this but I don't know how define a homeomorphism between X/~ and $T^2$ – Sophie Germain Mar 13 '13 at 22:18

Define a map $f$ from $I^2$ to $X$ which sends $(x,y)$ to $(x+1)\cdot (\textrm{cos}(2\pi yi),\ \textrm{sin}(2\pi yi))$. To make $I^2$ into a torus you have to glue opposite sides together, i.e. $(x,0)\sim_T(x,1)$ and $(0,y)\sim_T(1,y)$ for $x,y\in I$. The torus $T^2$ is then the quotient space $I^2/\sim_T$. Show that points are identified in $I^2$ if and only if their images are identified in $X$. Thus $f$ induces an injection $\hat f$ between $T^2$ and $X/\sim$. Since $f$ is surjective, $\hat f$ is bijective. Now you have to prove that $f$ is an identification mapping, that is a set $U\subseteq X$ is open iff its preimage $f^{-1}(U)$ is open. Finally, you should remember how this implies the openness of $\hat f$.

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