How to embbed $S^1\times [0, 1]$ in $\mathbb R^2$? How to define an embedding of $S^1\times [0, 1]$ in $\mathbb R^2$?
I tried to write $S^1=[0, 1]\sim$ where $0\sim 1$ and I defined $f:S^1\times [0, 1]\longrightarrow \mathbb R^2$ setting, $$f(x, t)=(\cos(\pi x+\pi/2), t),$$ then $f$ is well defined and it is continuous but it's not injective. When I define something injective it is not well defined, so I'm losing the hopes to find it.
 A: Consider, with $x \in \Bbb{S}^1$ and $t \in [0,1]$ 
$$
f(x,t) = \left( (t+1) \cos(2\pi x) ,  (t+1) \cos(2\pi x) \right)
$$
with $ 1 \leq t \leq 2$.
This $f$ is continuous and maps $\Bbb{S}^1 \times [0,1]$ onto a subset of $\Bbb{R}^2$ shaped like an annulus. $f$ is injective:  It is the projection  of a cylinder 
of unit radiaus and unit height, from a point along the cylinder axis one unit above the cylinder, onto a plane  one unit below the cylinder, and no rays from that point intersect the cylinder more than once.  So $f$ is an example of the desired embedding.
$f$ is not a surjection of $\Bbb{S}^1 \times [0,1]$ onto $\Bbb{R}^2$, and hence is not bijective. You can try to achieve a bijection by composing with a map of $[0,1]$ onto the half-line $p : p>0$.  For example, consider 
$$
g(x,t) = \left( \tan(\pi t/2) \cos(2\pi x) ,  \tan(\pi t/2) \cos(2\pi x) \right)
$$
But that is no longer injective, as the point $0,0$ in $\Bbb{R}^2$ is mapped to by the entire circle $(x \in [0,2\pi),t=0$ in the space $\Bbb{S}^1 \times [0,1]$.  In fact, I think there cannot be such an embedding that is a one-to-one, since the cylinder and the plane are not topologically equivalent.
