Surfaces constructed from curves: cylinders

We have defined some types of surfaces from curves, for example, generalized cylinders. Let $\alpha:I\rightarrow{\mathbb R}^3$ a curve contained in a plane $P$, which we suppose it is the plane $z=0$ and let $a\in {\mathbb R}^3$ be a vector that is not contained in $P$. The cylinder on base $\alpha$ in the direction of $a$ is the set $$S=\{\alpha(s)+ta:s\in I,t\in{\mathbb R}\}.$$ Of course, the parametrization is $$X:I\times{\mathbb R}\rightarrow{\mathbb R}^3, X(s,t)=\alpha(s)+ta.$$ If we prove that $S$ is a surface, it is immediate that $X$ is differentiable, $X_s=\alpha'(s)$, $X_t=a$ and both vectors are independent linearly. The difficulty appears when we want to prove that $X$ is a parametrization. The sets $I\times{\mathbb R}$ and $S$ are open in ${\mathbb R}^2$ and $S$, respectively. Also, it is immediate that $X$ is continuous. It remains to prove that $X$ is biyective and $X^{-1}$ is continuous. Of course, if $\alpha$ is not one-to-one, then $X$ is not, as in the next pictures (here the vector $a$ is $a=(1,1,1)$.

1. For this reason, we suppose two cases: $\alpha:I\rightarrow{\mathbb R}^3$ is an embedding or 
2. $\alpha:{\mathbb R}\rightarrow {\mathbb R}^3$ is a simple closed curve.

In the first case, $X$ is one-to-one. If $(x,y,z)=(\alpha_1(s)+ta_1,\alpha_2(s)+t a_2,ta_3)$, then $t=z/a_3$ and so $$s=\alpha^{-1}(x-\frac{z}{a_3} a_1,y-\frac{z}{a_3} a_2).$$ It is immediate that $X^{-1}$ is continuous.

In the second case, $\alpha$ is an embedding in an interval of length less than $T$, where $T>0$ is the period of $\alpha$. 

We have the next pictures for the simple closed curve $\alpha(s)=(3 \cos (s),\sin(s)+\cos(s)+\cos(2s)$ and $a=(0,0,1)$.