What is the boundary of the solid torus $\mathbb D^2 \times \mathbb S^1$? And extend it to higher dimension. We know that the boundary of the 3-manifold solid torus $\mathbb D^2 \times \mathbb S^1$ is $\mathbb S^1 \times \mathbb S^1$.
Now the question is as follows:


How does this generalize to dimensions larger than 3?


Can someone please let me know what does this question means? Does it say that extend it for higher genus?
Thanks!
 A: No, higher dimension is not the same thing as higher genus. The genus of a surface refers to how many "handles" it has, while the dimension is for which $n$ is it locally homeomorphic to $\mathbb{R}^n$ (and do not even use the word "surface" unless $n=2$.)
The solid torus $\mathbb{D}^2\times \mathbb{S}^1$ is a 3-dimensional manifold, and its boundary the torus $\mathbb{S}^1\times \mathbb{S}^1$ is 2-dimensional.
A higher dimensional analogue of this question would be "what is the boundary of $\mathbb{D}^3\times \mathbb{S}^2$? And the answer would be $\mathbb{S}^2\times \mathbb{S}^2.$ 
Or maybe just one dimension higher, "what is the boundary of $\mathbb{D}^3\times \mathbb{S}^1$? Answer: $\mathbb{S}^2\times \mathbb{S}^1.$
Or what is the boundary of $\mathbb{D}^2\times (\mathbb{S}^1)^{n-1}$? Answer: $\mathbb{S}^1)^n$, sometimes called the $n$-dimensional torus or $n$-torus (but don't confuse it with the $n$-handled torus).
More generally the boundary of $\mathbb{D}^{m+1}\times \mathbb{S}^n$ is $\mathbb{S}^{m}\times \mathbb{S}^n.$
Even more generally, we have for any manifolds with boundary $X$ and $Y$,
$$\partial(X\times Y)=\partial X\times Y\cup X\times\partial Y.$$
