Whether $\mathbb{T}\times [0,1]$ is diffeomorphic to $\mathbb{D}^2$?

Question

Here $\mathbb{T}$ denotes the torus $\mathbb{R}\backslash \mathbb{Z}$, and $\mathbb{D}^2$ the closed unit ball in $\mathbb{R}^2$. Since $\mathbb{T}\times [0,1]$ and $\mathbb{D}^2$ are both smooth compact manifolds with boundary (I'm not sure about this statement, if I was wrong, please correct me), can we establish a diffeomorphism between them? More precisely, can we find a smooth map $\varphi:\mathbb{T}\times [0,1]\to \mathbb{D}^2$ which admits a smooth inverse $\varphi^{-1}:\mathbb{D}^2\to \mathbb{T}\times [0,1]$?

Answer 1

This isn't possible. The 1-torus $\mathbb{T}$ is diffeomorphic to $S^1$, hence $\mathbb{T}\times[0,1]$ is homotopic to $S^1$ (it's a cylinder). So, $\mathbb{T}\times[0,1]$ can't be diffeomorphic to $\mathbb{D}^2$... they aren't even homotopic (for instance, they have different fundamental groups).

Answer 2

One can use algebraic topology to prove that the spaces are not even homotopy equivelent (for example, they have non-isomorphic fundamental groups).

Here is an alternative approach using only the smooth structures. We work with manifolds with boundary. Each diffeomorphism $f : M \to N$ between such manifolds restricts to a diffeomorphism $\partial f: \partial M \to \partial N$ between their boundaries. For our purposes it even suffices to know that $ f(\partial M) \subset \partial N$.

Assume that there exists a diffeomorphim $f : \mathbb{D}^2 \to \mathbb{T} \times [0,1]$. We have $\partial \mathbb{D}^2 = \mathbb{T}$ which is connected and $\partial \mathbb{T} = \mathbb{T} \times \{0,1\}$ which has two connected components.

Since $f(\mathbb{T}) \subset \mathbb{T}^2 \times \{0,1\}$, we must have $f(\mathbb{T}) \subset \mathbb{T} \times \{i\}$ either for $i = 0$ or $i=1$. Hence $\mathbb{T} \times \{0,1\} = f(f^{-1}(\mathbb{T} \times \{0,1\})) \subset f(\mathbb{T}) \subset \mathbb{T} \times \{i\}$, a contradicton.