How to determine a quotient space. $\mathbb{R} \times \mathbb{R}/\sim$, where $(x,y) \sim (x',y') \text{ iff } x+y'=x'+y.$

Question

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.

Answer 1

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 $1$.

Answer 2

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:

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

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