Is there a continuous function $f\colon\mathbb{R}\to\mathbb{R}^2$ such that the quotient space $\mathbb{R}/f$ is not embeddable into $\mathbb{R}^2$? Given a continuous function $f\colon X\to Y$, denote by $X/f$ the quotient of $X$ with respect to the equivalence relation $x\sim y\,\Leftrightarrow\,f(x)=f(y)$.
That is, the elements of $X/f$ are the equivalence classes and a set $A$ of equivalence classes is open in $X/f$ if and only if $\bigcup A$ is an open subset of $X$.
The mapping $\hat f\colon X/f\to f[X]$ defined by $\hat f([x])=f(x)$ is a bijection and the continuity of $f$ implies that $\hat f$ is continuous as well.
The mapping $f\colon X\to f[X]$ is called a quotient mapping whenever $\hat f$ is a homeomorphism, hence an embedding of $X/f$ into $Y$.
It can be proved that for any continuous function $f\colon\mathbb{R}\to\mathbb{R}$, $\hat f$ is actually a homeomorphism. The proof is based on Intermediate Value Theorem (see here).
Let us consider a non-intersecting curve $f\colon\mathbb{R}\to\mathbb{R}^2$ with a non-trivial limit (that is, $y_n\to y$ for some $y_n,y\in f[\mathbb{R}]$, but $x_n\nrightarrow x$ for any $x_n,x$ such that $f(x)=y$ and $f(x_n)=y_n$, for every $n$).
Then $\hat f$ is not a homeomorphism.
But in this case, $\mathbb{R}/f$ is clearly embeddable into $\mathbb{R}^2$ since $\mathbb{R}/f$ is homeomorphic to $\mathbb{R}$.
My question is, does there exist a continuous function $f\colon\mathbb{R}\to\mathbb{R}^2$ such that $\mathbb{R}/f$ is not embeddable into $\mathbb{R}^2$?
As it was mentioned above, there always exists a one-to-one continuous mapping from $\mathbb{R}/f$ into $\mathbb{R}^2$, but I am looking for the case when no such mapping is open.
 A: Consider $f:\Bbb R\to \Bbb R^2$ a "natural" continuous parametrization of the Hawaiian earring: say, $$f(x)=\begin{cases} (0,0)&\text{if }x\le 1\\ \frac{1}{\lfloor x\rfloor}\left(1+\cos(\pi(-1+2x-2\lfloor x\rfloor)),\sin(\pi(-1+2x-2\lfloor x\rfloor))\right)&\text{if }x>1\end{cases}$$
$\Bbb R/f$ is the space obtained from $[1,\infty)$ by collapsing $\Bbb N$ to a single point. Alternatively, it can be described as a bouquet of infinite-countably many circles. Either way, it cannot be realized as a subspace of $\Bbb R^2$, because it is not first-countable.
A: I think I understand you problem.  Assuming that, ...
(This is a classic example of an immersion that is not an embedding.  It appears as the second image at Immersion (English Wikipedia).  I haven't seen this particular explicit realization of it, but surely someone has produced this example before.)  Let $n(t) = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-t^2/2}$, that is, $n$ is the PDF of the standard normal distribution.  Then let \begin{align*}
    f:\mathbb{R} &\longrightarrow \mathbb{R}^2  \\
    t &\longmapsto (n'(t), n^{(3)}(t))  \text{,}
\end{align*}
where the superscript $(3)$ indicates taking the third derivative.  The embedding is a (mildly distorted) figure "$8$" where $0 \mapsto (0,0)$, the crossing of the "$8$" and also $\lim_{t \rightarrow -\infty} f(t) = \lim_{t \rightarrow \infty} f(t) = (0,0)$.  That is, the open ends of the domain $\mathbb{R}$ are mapped to open ends of the curve pressed up against $(0,0)$ transversely to the curve in a neighborhood of $t= 0$.  Consequence: embedding fails at $(0,0)$ because there are no "$\times$"-shaped open sets of $\mathbb{R}$.  (One could also observe that the domain is contractible but the image has nontrivial fundamental group, $\mathbb{Z} \oplus \mathbb{Z}$.  Since fundamental groups are homeomorphism invariants, this is not an embedding.)
A plot of this $f$ for $t \in [-3.5,3.5]$:

One might try to go from here to the "simpler" $t \mapsto \left( \cos(2 \tan^{-1} t), \sin(2 \tan^{-1} t) \right)$, but this doesn't close the loop, so the image is contractible.  You really need that limit point to be covered by some other part of the curve.
