Why is $S^1 \times [0, 1]$ an annulus?

Question

($S^1$ is the unit circle.) $S^1 \times [0, 1]$ looks like a closed disk with no holes in it to me, but an annulus has a hole in the center of a circle. If so, then what is the radius of the "hole" for the annulus $S^1 \times [0, 1]$?

Answer 1

You're a little too hung-up on the 'unit circle' representation of $S^1$ and not thinking abstractly enough. If you want to think concretely, consider $S^1$ as the circle $x^2+y^2=1$ and take the product with the interval $0\leq z\leq 1$. In particular, you shouldn't be identifying the points $\{(s,0): s\in S^1\}$, which is what you're implicitly doing.

Another useful check: since the interval $[0,1]$ is isomorphic to the interval $[1,2]$ then certainly $S^1\times[0,1]$ should be isomorphic to $S^1\times[1,2]$, and the latter is (hopefully) more obviously an annulus any way you look at it.

Now, can you find an isomorphism from the 2d annulus $\{(r,\theta): 1\leq r\leq 2\}$ (in polar coordinates) to the first 'three-dimensional' representation given above?

Answer 2

Here's an explicit correspondence: $$ (\theta, t) \mapsto ((3+t) \cos \theta, (3 + t) \sin \theta) $$ This shows that $S^1 \times [0, 1]$ is diffeomorphic to the annulus in the plane whose center is the origin, with inner radius 3 and outer radius 4. You can probably adjust the mapping to make this work with whatever inner and outer radii you like.

If you want to think of $S^1$ as the set of points $(x, y)$ with $x^2 + y^2 = 1$, then a slightly different map is

$$ (x, y, t) \mapsto \left(\frac{1+t}{2} x, \frac{1+t}{2} y\right) $$ The image of that map is an annulus of inner radius $1/2$, corresponding to $t = 0 \in [0, 1]$ and outer radius $1$, corresponding to $t = 1 \in [0, 1]$.