Quotient of $\mathbb{R}^n$ with an unbounded equivalence class homeomorphic to $\mathbb{R}^n$?

Question

Let $X=\mathbb{R}^n$. Suppose $X$ has an equivalence relation $\sim$ such that at least one class consists of a line (a $1$-D subspace) through the origin. If $X^*=X/{\sim}$, is it possible for $X$ to be homeomorphic to $X^*$?

My intuition says no since $X^*$ pinches an infinitely long line to a point, and it's well-known a line and a point are not homeomorphic, though this doesn't truly apply here since we're in higher dimensions, so the cardinality argument doesn't apply. Wikipedia has a nice long list of topological invariants, though none of them immediately popped out at me as giving an answer to the question (apart from the $n=1$ case as pointed out). Once invariant I did come across was 'topological homogeneity' but it seems at the outset that proving $X^*$ is not topologically homogeneous in general is no small task (or even convincing myself it's true).

I haven't been able to find an example thus far, but it has been some time since I took topology. Any help would be greatly appreciated.

Answer 1

Let me elaborate on Daron's answer and show more generally that for any $n$, there exists a quotient map $\mathbb{R}^n\to\mathbb{R}^n$ such that one of the equivalence classes contains a hyperplane. This is trivial for $n\leq1$, so I will assume $n>1$. First, let me give a criterion for a map to be a quotient map.

Lemma: Let $Y$ be a compactly generated Hausdorff space, and let $f:X\to Y$ be a continuous surjection such for each compact $K\subseteq Y$, there is a compact $L\subseteq X$ such that $K\subseteq f(L)$. Then $f$ is a quotient map.

Proof: Suppose $U\subseteq Y$ is such that $f^{-1}(U)$ is open. Since $Y$ is compactly generated, it suffices to show $U\cap K$ is open in $K$ for any compact $K\subseteq Y$. Choose a compact set $L\subseteq X$ such that $K\subseteq f(L)$. Note that since $L$ is compact and $Y$ is Hausdorff, the restriction of $f$ to $L\to f(L)$ is a quotient map. Now $f|_L^{-1}(U\cap f(L))=f^{-1}(U)\cap L$ is open in $L$, and hence $U\cap f(L)$ is open in $f(L)$. It follows that $U\cap K$ is open in $K$, as desired.

Now let us construct a quotient map $\mathbb{R}^n\to\mathbb{R}^n$. First, let $g:\mathbb{R}^{n-1}\to S^{n-1}$ be a continuous surjection which moreover is still surjective when restricted to some compact subset $A\subset\mathbb{R}^{n-1}$ (it is easy to see this is possible for $n>1$). Define $f:\mathbb{R}^n\to\mathbb{R}^n$ by $f(t,x)=tg(x)$ for $t\in\mathbb{R}$ and $x\in\mathbb{R}^{n-1}$. Then $f$ is continuous and surjective, and it is easy to see it satisfies the hypotheses of the lemma (any compact subset of $\mathbb{R}^n$ is contained in $f([0,N]\times A)$ for $N$ sufficiently large). Thus $f$ is a quotient map. Moreover, $f$ is constant on the hyperplane $t=0$.

Answer 2

I thunk the answer is yes for $n=2$, Define a continuous map $f : \mathbb R^2 \to \mathbb R^2$ to take each vertical line of constant positive $x$-value $r$ and wrap it around the circle of radius $r$ with rightmost point at the origin. There are several ways to do this. Note then the $y$-axis is mapped to the origin. Define the function for the left half-plane by symmetry. Then use the equivalence relation $x\sim y \iff f(x)=f(y)$. Can you prove that $f$ is continuous? We need this to make the quotient space equal to the image.