Homeomorphism between a quotient topology on $\mathbb{R}^2$ and $\mathbb{R}$ Consider $X = \mathbb{R}^2$ endowed with the standard topology. Define the equivalence relation $\sim$ by $(x_1,y_1) \sim (x_2,y_2)$ if $x_1 = x_2$. Let $\tau$ be the quotient topology on $X/\sim$. Show that $(X/\sim, \tau)$ is homeomorphic to $\mathbb{R}$ with the standard topology.
It looks like the equivalence classes are $E_x = \{x\} \times Y$. So I think I want to show that $\{x\} \times Y$ is homeomorphic to $Y$. I noticed that the fact is stated (without proof) in Munkres ch. 26, so it must be very obvious. But I'm not sure how to go about showing it. Any help would be appreciated.
 A: HINT:
What you are looking for is a continuous bijection between $X/\sim$ and $\mathbb{R}$ which has continuous inverse. You note that the equivalence classes in $X/\sim$ look like $\{x\} \times Y$.
So, for each element $x$ in $\mathbb{R}$ you have one equivalence class: $\{ x\} \times Y$. I.e. you have a map from $\mathbb{R}$ into $X/\sim$. Is this continuous? Specifically: What do open sets in $X/\sim$ look like? Are their pre-images open?
Perhaps there exists an inverse map. Is that continuous? If so, you have a continuous map with a continuous inverse, and you have the homeomorphism you want.
A: Every value of $x$ defines (labels) an equivalence class and the quotient space is the set of equivalence classes: hence, it is the set of all $[x]$, that is the real line. Now you have to show by using the definition of quotient topology that also the induced topology is the same of the standard real line. You have to consider the sets of the plane with an open preimage under the surjective map $(x,y) \rightarrow x$.
