$(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic Is picture the only way to show that $(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic? It is obvious when we think about a cylinder. However, is there another clever way to show this more rigorously?
Here, $D^n$ is the closed unit disk in $\Bbb R^n$.
 A: We begin with some preparations. Let $\lVert - \rVert_1, \lVert - \rVert_2$ be two norms on $\mathbb R^m$. Since all norms are equivalent, both induce the usual Euclidean topology on $\mathbb R^m$. Both $\lVert - \rVert_i$ are continuous functions $\mathbb R^m \to  \mathbb R$.
Let $B^m(\lVert - \rVert_i) = \{ x \in \mathbb R^m \mid \lVert x \rVert_i \le 1 \}$ be the closed unit $\lVert - \rVert_i$-ball in $\mathbb R^m$. Define
$$h_{12} : B^m(\lVert - \rVert_1) \to  B^m(\lVert - \rVert_2), h(x) = \begin{cases} \frac{\lVert x \rVert_1}{\lVert x \rVert_2}x & x \ne 0 \\ 0 & x = 0 \end{cases}$$
This is well-defined because if $x \in B^m(\lVert - \rVert_1), x \ne 0$, then $\lVert x \rVert_1 \le 1$ and hence $\lVert h_{12}(x) \rVert_2 = \lVert x \rVert_1 \le 1$. It is continuous in all $x \ne 0$. To verify continuity in $0$, it suffices to observe that $\lVert h_{12}(x) - h_{12}(0) \rVert_2 = \lVert h_{12}(x) \rVert_2 = \lVert x \rVert_1 \to 0$ as $\lVert x \rVert_1 \to 0$. Recall that all norms are equivalent.
We have $h_{21} \circ h_{12} = id$ and $h_{12} \circ h_{21} = id$, thus $h_{12}$ is a homeomorphism. Geometrically $h_{12}$ linearly distorts each line segment $L^1_b$ connecting $0$ with a boundary point $b$ of $B^m(\lVert - \rVert_1)$, i.e. $\lVert b \rVert_1 = 1$, to the line segment $L^2_b$ connecting $0$ with the  boundary point $\frac{b}{\lVert b \rVert_2}$ of $B^m(\lVert - \rVert_2)$.
Let $B^m_+(\lVert - \rVert_i) = \{ x = (x_1,\ldots,x_m) \mid \lVert x \rVert_i \le 1 , x_m \ge 0  \}$ and $B^m_-(\lVert - \rVert_i) = \{ x = (x_1,\ldots,x_m) \mid \lVert x \rVert_i \le 1 , x_m \le 0  \}$ be the upper and lower closed unit $\lVert - \rVert_i$-ball in $\mathbb R^m$, respectively.
Because $h_{12}$ does not change the sign of any coordinate, it restricts to homeomorphisms
$$h^\pm_{12} : B^m_\pm(\lVert - \rVert_1) \to B^m_\pm(\lVert - \rVert_2) .$$
For the following constructions drawing pictures will be useful.
On $\mathbb R^{n+1} = \mathbb R^n \times \mathbb R$ we have the usual Euclidean norm $\lVert - \rVert$ and the norm $\lVert (\xi,t) \rVert' = \max(\lVert \xi \rVert, \lvert t \rvert)$. Here $(\xi,t ) \in \mathbb R^n \times \mathbb R$.
Then $B^{n+1}_\pm(\lVert - \rVert) =  D^{n+1}_\pm$ = upper/lower closed unit $\lVert - \rVert$-ball in $\mathbb R^{n+1}$ and $B^{n+1}_+(\lVert - \rVert') = D^n \times I, B^{n+1}_-(\lVert - \rVert') = D^n \times [-1,0]$. As above we get homeomorphisms
$H_+ : D^n \times I \to D^{n+1}_+$ and $H_- : D^n \times [-1,0] \to D^{n+1}_+$. Define $s :  D^n \times I \to D^n \times [-1,0], s(\xi,t) = (\xi,t-1)$. This is a homeomorphism. Let $H'_- = H_- \circ s$. By construction we have $H_+(D^n \times \{0\}) =  D^n \times \{0\}$ and $H'_-(D^n \times \{0\} \cup S^{n-1} \times I) = S^n_- =  \{ x = (x_1,\ldots,x_{n+1}) \mid \lVert x \rVert = 1, x_{n+1} \le 0  \}$. Now define
$$\phi : D^{n+1}_+ \to D^{n+1}_-, \phi(\xi,t) = (\xi,t - \sqrt{1 - \lVert \xi \rVert^2}) .$$
This a well-defined continuous map. Note $\lVert (\xi,t - \sqrt{1 - \lVert \xi \rVert^2}) \rVert^2 = \lVert \xi \rVert^2 - 2t\sqrt{1 - \lVert \xi \rVert^2} + 1 - \lVert \xi \rVert^2 = 1 - 2t\sqrt{1 - \lVert \xi \rVert^2} \le 1$ since $t \ge 0$ and $t - \sqrt{1 - \lVert \xi \rVert^2} \le 0$ since $t \le \sqrt{1 - \lVert \xi \rVert^2}$. But $\phi$ is a homeomorphism (its inverse is given by $(\xi,t) \mapsto (\xi,t + \sqrt{1 - \lVert \xi \rVert^2})$). Geometrically $\phi$ moves each line segment $L_\xi$ connecting $(\xi,0)$ with $(\xi,\sqrt{1 - \lVert \xi \rVert^2}$ down to the line segment $L'_\xi$ connecting $(\xi,-\sqrt{1 - \lVert \xi \rVert^2}$ with $(\xi,0)$.
By comnstruction we have $\phi( D^n \times \{0\})  =  S^n_-$. This answers your question.
