$(0,1)\to\mathbb{R}^2$ injective, continuous, not a homeomorphism on the image

Question

Consider the map $$\gamma\colon (0,1)\to\mathbb{R}^2,\ t\mapsto (\cos(2\pi t),\sin(2\pi t)).$$ This is an example of a map which is continuous and injective but not a homeomorphism onto the image, since the inverse could not be continuous. In fact, two points arbitrarily close to each other in a small neighbourhood of $(1,0)$ would go far apart in the preimage. By definition, a function is continuous if the preimage of every open set is open in the domain. How could I find an open set in the support of this curve which is sent to a non-open set in the interval?

Answer 1

The given map is a homeomorphism onto the image: Let $I:=(0,1)$, $S:=\gamma(I)$, and consider a point $z\in S$. Then $t:=\gamma^{-1}(z)\in I$. Any neighborhood $V$ of $t$ contains an open interval $(a,b)$ such that $0<a<t<b<1$. The set $$\Omega:=\{(x,y)\in{\mathbb R}^2\ |\ x^2+y^2>0, \ 2\pi a<\arg(x,y)<2\pi b\}$$ is open in ${\mathbb R}^2$, whence $U:=\Omega\cap S$ is an open subset of $S$ which contains the point $z$. Therefore $U$ is an open neighborhood of $z$, and $\gamma^{-1}(U)=(a,b)\subset V$.

Here is an example of a continuous injective map $f:\ I\to {\mathbb R}^2$ which is not a homeomorphism onto its image: $$f(t)\ :=\ \cases{(6t-1,0) & $\bigl(0 < t\leq{1\over3}\bigr)$\cr (2-3t, 3t-1) & $\bigl({1\over3}\leq t\leq{2\over3}\bigr)$ \cr (0,3-3t) & $\bigl({2\over3}\leq t < 1\bigr)$ \cr}\quad.$$ Drawing a figure one sees that the inverse map $f^{-1}$ is not continuous at $(0,0)=f\bigl({1\over6}\bigr)$.