Let $\gamma (t) : \mathbb R \to \mathbb R^2$ be the function $$\gamma(t)=\left(\frac{(1+t^2)t}{1+t^4},\frac{(1-t^2)t}{1+t^4}\right)$$
- Prove that the function is $\gamma$ is differentiable, regular and simple.
- Determine $\lim_{t \to -\infty} \gamma(t)$ and $\lim_{t \to +\infty} \gamma(t)$ and deduce that $\gamma$ is not a homeomorphism between $\mathbb R$ and the lemniscate.
I could show that $\gamma$ is differentiable and regular (i.e., $\gamma'(t) \neq 0$). I am having some difficulty showing it is simple. Suppose there are $t_1,t_2$ such that $\gamma(t_1)=\gamma(t_2)$. Then we have $$(1) \space \frac{(1+{t_1}^2)t}{1+{t_1}^4}=\frac{(1+{t_2}^2)t}{1+{t_2}^4},$$$$(2) \space\frac{(1-{t_1}^2)t}{1+{t_1}^4}=\frac{(1-{t_2}^2)t}{1+{t_2}^4}$$ By analyzing appart the case $\gamma_2(t)=0$, we can divide $(1)/(2)$ and get to the equation $$\frac{1+{t_1}^2}{1-{t_1}^2}=\frac{1+{t_2}^2}{1-{t_2}^2}$$
I couldn't conclude from here that $t_1=t_2$
I've also calculated $\lim_{t \to -\infty} \gamma(t)=0=\lim_{t \to +\infty}$, but I have no idea how to deduce from here that $\gamma$ is not a homeomorphism.
The geometric symmetry of the lemniscate $L$ is reflected in algebraic symmetries of the parametrization $t\mapsto\gamma(t)$. These can be used to simplify the proof of injectivity, as follows:
From $\gamma(-t)=-\gamma(t)$ and $x(t)>0$ when $t>0$ it follows that it is enough to prove injectivity on ${\mathbb R}_{>0}$. Furthermore we easily check that $$x\left({1\over t}\right)=x(t),\quad y\left({1\over t}\right)=-y(t)\ne0\qquad(t>0)\ .$$ Therefore it is enough to verify that $t\mapsto x(t)$ is strictly increasing on $[0,1]$. The latter follows immediately from $$x'(t)={(1-t^2)(1+4t^2+t^4)\over(1+t^2)^4}\ ,$$ which you have computed anyway.
Concerning the last part: The points $${\bf z}_n:=\gamma(n)\in L\qquad(n\in{\mathbb N})$$ converge to the point ${\bf 0}\in L$, but their inverse images $\gamma^{-1}({\bf z}_n)=n$ form a divergent sequence on ${\mathbb R}$. It follows that $\gamma:\>{\mathbb R}\to L$ cannot be a homeomorphism.