Function on "figure eight" which is not a diffeomorphism due to its topology

Question

In my lecture, we've considered $G(x)=F(g(x))$ the "figure eight", where $F(x) = (2\cos(x-\frac{1}{2}\pi), \sin(2(x-\frac{1}{2}\pi))$ and $g(x)=\pi+2\arctan(x)$.
I understand that this is an example for a bijective immersion which is not a homeomorphism since the figure eight and $\mathbb{R}$ are not homeomorphic.
Now, I should give an example of a function which maps from figure eight to figure eight bijectivly and homeomorphically, but is not a diffeomorphism due to its topology.
My idea was to consider the reflection of the figure eight (let's call $G(\mathbb{R}))=A$) an the x-axis, i.e. $R:A\rightarrow A, R(x,y)=(x,-y)$. Intuitively, I thought that this could work since we are able to approach $0$ in A with a sequence $(x_n)_{n \in \mathbb{N}}\subset \mathbb{R}$ s.t. $x_n \rightarrow 0$ but also with $x_n \rightarrow \pm \infty$.
Is my idea correct and if yes, could you help me proving this?
Thank you!