What is boundary of $S^1\times I^2$

Question

As I know, the general cylinder is $S^k \times R^l$. For talking the boundary of general cylinder, let $I=[0,1]$. Consider $S^k \times I^l$, obviously, when $l=1$, the boundary of $S^k \times I^l$ is two unit spheres $S^k$. But when $l\ge 2$, it is hard for me to image it. For simple, how to know what is the boundary of $S^1\times I^2$ ?

Answer 1

In general, the boundary of the products of two manifolds $M\times N$ is $\partial M\times N\cup M\times\partial N$.

Now $\partial S^k=\varnothing$ and $\partial I^\ell = S^{\ell-1},$ (since a filled hypercube $I^\ell$ is homeomorphic to the disk $D^\ell$, its boundary is a hypercube homeomorphic to a sphere), so for a sanity check, let's look at $$\partial(S^k\times I^1)=\partial(S^k)\times I^1\cup S^k\times \partial(I^1) = \varnothing\times I^1\cup S^k\times S^0 = S^k\times \{-1,1\}=S^k\coprod S^k,$$ so yes, that's two copies of $S^k$ since $\partial(I^1)=S^0$ is two points, and the boundary of a sphere is empty.

Next for $k=1,\ell=2,$ the formula gives us $\partial (S^1\times I^2) = S^1\times S^1$, a torus, which checks out (note that the second $S^1$ here is actually a square. The boundary of a squared off solid torus is a squared off torus).

For your general case then, $\partial (S^k\times I^\ell) = S^k\times S^{\ell-1}.$