Continuous, injective curve that is not an embedding (Descartes folium)

Question

Consider the curve $x:\left(-1,+\infty\right) \to \mathbb{R}^2$ given by:

$x\left(t\right)= (x_1(t),x_2(t)) = \Big(\frac{3t}{1+t^3},\frac{3t^2}{1+t^3}\Big)$

i.e. "half" the parametrisation of Descartes folium

Then it is continuous and injective, but is not a homeomorphism onto its image.

Why?

Injectivity is ok.

For the second point, the inverse function is:

$x^{-1}(t) = \begin{cases} \frac{x_2}{x_1} & \text{ if }x_1 \in \left(-\infty,a\right]\setminus\{0\},a>0,x_2 \in \left[0,+\infty\right) \\ 0 & \text{ if }x_1=0 \end{cases}$

Now, where the problem for continuity of the inverse occurs?

Answer 1

Consider $f:\mathbb{R}^2\supset C \to [-1,+\infty)$ where $C$ is image of $x$, and $f=x^{-1}$: $$ f\left((x_1,x_2)\right) = \left\{ \begin{array}{lr} 0 & : x_1=0\\ \frac{x_2}{x_1} & : otherwise \end{array} \right. $$ Define sequence $\{s_i\}_{i=1}^{\infty}$ where $s_k=(\frac{3k}{1+k^3},\frac{3k^2}{1+k^3})$. $$\{s_i\}_{i=1}^{\infty}\subset C$$ $$f(0,0)=0$$ $$f(s_k)=k$$ $$\lim_{k\to +\infty}s_k=(0,0) $$ $$\lim_{k\to +\infty}f(s_k)=+\infty$$ $$\lim_{s_k\to (0,0)}f(s_k)\neq f(0,0)$$

therefore $f$ is not continuous.

Answer 2

So, $x(0) = (0,0)$. Further, note that $\lim_{t\to\infty} x(t) = (0,0)$. So, for any $\varepsilon > 0$ there exists some arbitrarily large $t_0$ (say, $t_0 > 1$) such that $\vert x(t_0)\vert < \varepsilon$. Therefore, the preimage of any $\varepsilon $-ball containing $(0,0)$is not going to be contained within a $\delta$-ball in $\mathbb{R}$ with radius less than $1$, and so the inverse is not continuous.