How to determine a quotient space. $\mathbb{R} \times \mathbb{R}/\sim$, where $(x,y) \sim (x',y') \text{ iff } x+y'=x'+y.$ On $\mathbb{R}\times \mathbb{R}$ let $\sim$ be an equivalence relation defined as follows: 
$$(x,y) \sim (x',y') \text{ iff } x+y'=x'+y.$$
Prove $\sim$ is an equivalence relation, determine equivalence class for $\hat{(2,5)}$ and determine $\mathbb{R}\times \mathbb{R}/\sim$.
It was easy to prove that $\sim$ is an equivalence relation. to determine the equivalence class for $\hat{(2,5)}$ 
I did as follows:
$\hat{(2,5)}=\{(x,y):(x,y)\sim (2,5)\}.$ 
So, $x+5=y+2$, $y=x+3$. we can conclude that this equivalence class is  the line $y=x+3$. 
Any help to find out the $\mathbb{R}\times \mathbb{R}/\sim$ it will be appreciated. 
 A: It's similar to the case of the equivalence class of $(2,5)$: the equivalence class of $(a,b)$ is$$\{(x,y)\in\mathbb R^2\mid y-x=b-a\}.$$So, it's the line with slope $1$ passing through $(a,b)$. And therefore $\mathbb R\times\mathbb R/\sim$ is the set of all lines in $\mathbb R^2$ whose slope is $1$.
A: Here is a general setup: you have  topological spaces $X$, $Y$ and a continuous surjective map from $X$ to $Y$
$$p\colon X \to Y$$
Consider the equivalence relation on $X$
$$x \simeq y \textrm{ if and only if } p(x) = p(y)$$
Let $\tilde X$ be the quotients space $X/\simeq$ with the quotient topology. We get a induced continuous bijection
$$\bar p \colon \tilde X \to Y$$
defined by
$$\bar p(\tilde x) = p(x)$$
Now $\bar p$ will be a homeomorphism in some cases:

*

*if $p$ is an open map, that is $p(U)$ open for every $U\subset X$ open ( this is your situation with $p\colon \mathbb{R}\times \mathbb{R} \to \mathbb{R}$, $p(x,y) = y-x$


*if $p$ is a closed map ( for example when $X$, $Y$ are both compact and Hausdorff)
