Question: Let $∼$ be the equivalence relation on $\mathbb{R}^2$ given by ($x_1,x_2)∼ (y_1,y_2)$ if and only if $x_2 = y_2$. Show that the identification space $\mathbb{R}^2/∼$ is homeomorphic to $\mathbb{R}$.
I have tried to define a function $f:\mathbb{R}^2/∼ \to \mathbb{R}$ by $f(x,y)=y$ for all $(x,y)\in\mathbb{R^2}$. Then $f$ is clearly a bijection.
My trouble is to show $f$ and $f^{-1}$ are continuous. I think $f$ is a projection map, so it is continuous, am I right?
Any help is appreciated.
Best Answer
Let $[y]$ denote the equivalence class of any point $(x,y) \in \mathbb{R}^2$, according to $\sim$. That is, $[y] = \{(x,y) : x \in \mathbb{R}\}$. Now let $U$ be an open subset of $\mathbb{R}$. Then $f^{-1}(U) = \{[y] : y \in U\}$ and this set is open in $\mathbb{R}^2/ \sim$ by the definition of quotient toplogy, since if $\pi : \mathbb{R}^2 \to \mathbb{R}^2 / \sim$ given by $\pi(x,y) = [y]$ is the quotient map, then $\pi^{-1}(f^{-1}(U)) = \mathbb{R}\times U$, which is open in $\mathbb{R}^2$.
That $f^{-1}$ is continuous means that if $W$ is open in $\mathbb{R}^2/ \sim$, then $f(W)$ is open in $\mathbb{R}$. So fix such $W$ and note that $\pi^{-1}(W) = \mathbb{R}\times W$ must be open in $\mathbb{R}^2$, so $W$ is open in $\mathbb{R}$. But $W = f(W)$, so we're done.