This question was left as an exercise in a seminar of topology and I was not able to fully solve the question at my home.
Question : $\mathbb{R}^2/ \mathbb{R}(1,1)$ equipped with quotient topology. Show that it is homeomorphic to $\mathbb{R}$.
Attempt: I defined a map $f:\mathbb{R}^2 \to \mathbb{R}$ by $f(x,y) =x-y$. It's kernel is $\mathbb{R} (1,1)$. So, I have an induced bijection from a quotient space to $\mathbb{R}$. This map is continuous by the definition of quotient topology.( How to show that inverse image of this map is continuous?) Now, I know that $f$ is the composition of the homeomorphism $g:\mathbb{R}^2 \to \mathbb{R}^2 $ , $g(x,y) = (x-y,x+y)$ with the projection on the first factor. As this projection is an open map, $f$ is open.
But I am not able to move forward from this to prove that inverse of $f$ is continuous.
Can you please help!
Call $\pi\colon \mathbb{R}^2\to \mathbb{R}^2/\mathbb{R}(1,1)$ the projection, and $\overline{f}\colon \mathbb{R}^2/\mathbb{R}(1,1)\to \mathbb{R}$ the induced map. Then $f=\overline{f}\circ \pi$. As you showed, $f$ is open. Let $U$ be an open set of $\mathbb{R}^2/\mathbb{R}(1,1)$. Then $\pi^{-1}(U)$ is also open, and then: $\overline{f}(U)=\overline{f}(\pi(\pi^{-1}(U))=f(\pi^{-1}(U))$ is open. This proves that $\overline{f}$ is an open continuous bijection.
Now, general fact: every open continuous bijection $\varphi\colon X\to Y$ is an homeomorphism. In fact, let $U\subseteq X$ open. Then $(\varphi^{-1})^{-1}(U)=\varphi(U)$ is open, since $\varphi$ is open. Then $\varphi^{-1}$ is continuous.