Why the torus has open subsets diffeomorphic to a cylinder? If I take the torus $T$ of revolution in $\mathbb{R}^3$ is very intuitive for me that there is an open subset of $T$ that is diffeomorphic to a cylinder, for example $T$ minus a circle on the $xz$ plane. But how can I prove this in a rigourous way? This problem appeared to me when I learn how to calculate the cohomology groups of the deRham complex of $T$, where is assumed that we can take such a type of open subsets, and we use those open sets in the Mayer-Vietoris Sequence. Can anyone give me a help? Thanks.

It depends on your definition of the torus. For instance, if the torus is $S^1 \times S^1$ then you might remove $\{x_0\} \times S^1$.
We know $(S^1 \times S^1) \setminus (\{x_0\} \times S^1)$ is the same thing as $(S^1 \setminus \{x_0\}) \times S^1$, but it's pretty easy to see that $S^1 \setminus \{x_0\}$ is diffeomorphic to $(0,1)$, which means our set is diffeomorphic to $(0,1) \times S^1$, a cylinder.

I hope this helps ^_^
