Is the Warsaw circle a path in plane?

Question

A circle $S^1$ is a path in plane $\mathbb{R}^2$ since there exists a mapping $f:[0,1]\to S^1$.

The Warsaw circle is not homotopy equivalent to the circle, however it is shape equivalent to it.

My question is: Can the Warsaw circle be seen as a path in $\mathbb{R}^2$, i.e. is there a mapping $g:[0,1]\to W$ ($W$ stands for Warsaw's circle)?

My attempt: let $g$ be defined such that $g[0,\frac{1}{4} ]\to\{ 0 \}\times [-1,1]$ and $g[\frac{1}{4},\frac{1}{2} ]\to\{ 0 \}\times [-1,0]$ are linear mappings such that $g(0)=(0,1), \; g(\frac{1}{4})=(0,-1) \text{ and } g(\frac{1}{2})=(0,0)$. Let $g:(\frac{1}{2},0)\to W\backslash \left\{ 0 \right\}\times [-1,1]$ be an appropriate continuous mapping and $g(1)=(0,0)$. Then $g$ is continuous.

Answer 1

Your question is imprecise. A subset of the plane is never a path (which is a map $[0,1] \to \mathbb R^2$). I think you ask whether a subset $X \subset \mathbb R^2$ is the image of a path $f : [0,1] \to \mathbb R^2$ with the property that $f(s) = f(t)$ iff $\{s,t\} = \{0,1\}$. A necessary condition is that $f([0,1]) = X$.

There exist many maps $[0,1] \to W$ (e.g. all constant maps), but none of them is surjective. To see this, assume that there exists a continuous surjection $f : [0,1] \to W$. We have $W = A_1\cup A_2\cup A_3 \cup A_4 $, where $A_1=\{(x,\sin\frac{\pi}{x}):0<x \le 1 \}$, $A_2=\{ (0,y):-1\le y \le 1\}$, $A_3=\{(x,1+\sqrt{x-x^2}):0\le x \le 1\}$, $A_4=\{(1,y):0\le y \le 1\}$. See To show that Warsaw circle is simply connected. Pick $t_n \in [0,1]$ such that $f(t_n) =(\frac{1}{n + \frac 1 2},\sin (n\pi + \frac \pi 2))$. The sequence $(t_n)$ has a subsequence $(t_{n_k})$ converging to some $t_* \in [0,1]$. But then $f(t_{n_k}) \to f(t_*)$ which means that $f(t_*) = (0,\pm 1)$. There exists $\epsilon > 0$ such that $ \lVert f(t) - f(t_*) \rVert < \frac 1 2$ for $\lvert t - t_* \rvert \le \epsilon$. Thus the interval $J = [0,1] \cap [t_* - \epsilon,t_* + \epsilon]$ is mapped by $f$ into $W' = W \setminus \{(\frac 1 2, \frac 3 2)\}$. This shows that there exists a path in $W'$ connecting $f(t_*) = (0,\pm 1)$ with some point $f(t_{n_k}) \in A_1$. Hence $W'$ must be path connected which is known to be false. This is a contradiction, thus our assumption was false.