Example from tom Dieck's Algebraic Topology: $(D^n,S^{n - 1})\times ([0,1],\{0\}) \cong (D^n,\varnothing)\times ([0,1],\{0\})$ For pairs a topological spaces, tom Dieck defines a product as follows:

$(X,A)\times (Y,B) = (X\times Y, (X\times B)\cup(A\times Y))$.

On page 36 of the book Algebraic Topology tom Dieck provides the following example:

The assignment $H\colon (x,t) \mapsto ((1/\alpha(x,t))(1 + t)x, 2 - \alpha(x,t))$ where $\alpha(x,t) = \max\{2||x||,2 - t\}$ yields a homeomorphism of pairs $(D^n,S^{n - 1})\times ([0,1],\{0\}) \cong (D^n,\varnothing)\times ([0,1],\{0\})$

Since $D^n\times [0,1]$ is compact and Hausdorff, to prove that $H$ is a homeomorphism it suffices to prove that it is continuous and bijective. However, I have no idea how to do so, the map just seems intimidating.
 A: We have $\alpha : D^n \times [0,1] \to [1,2]$.


*

*$H : D^n \times [0,1] \to D^n \times [0,1]$ is well-defined: Write $H(x,t) =(H_1(x,t),H_2(x,t))$ with $H_1(x,t) = \frac{1+t}{\alpha(x,t)}x$ and $H_2(x,t) = 2 - \alpha(x,t)$. Explicitly we have
$$H(x,t)  = \begin{cases} \left( \frac{1+t}{2 -t}x,t \right)  & 2\lVert x \rVert \le 2-t \\ \left( \frac{1+t}{2\lVert x \rVert}x , 2(1 - \lVert x \rVert) \right)& 2\lVert x \rVert \ge 2-t\end{cases}$$
Obviously $H_2(x,t) \in [0,1]$ and
$$\left\lVert H_2(x,t) \right\rVert = \begin{cases} \frac{1+t}{2 -t}\lVert x \rVert \le \frac{1+t}{2-t} \frac{2-t}{2} = \frac{1+t}{2} \le 1   & 2\lVert x \rVert \le 2-t \\ \frac{1+t}{2\lVert x \rVert}\lVert x \rVert = \frac{1+t}{2} \le 1 & 2\lVert x \rVert \ge 2-t\end{cases}$$

*$H$ is continous: Obvious.

*$H$ is injective: Let $H(x,t) = H(x',t')$. Then the second coordinate shows us $\alpha(x,t) = \alpha(x',t')$ and then the first coordinate gives $(1+t)x = (1+t')x'$. Thus, if $x = 0$ or $x' = 0$, then both $x = x' = 0$ and $2-t = \alpha(0,t) = \alpha(0,t') = 2-t'$, i.e. $t = t'$. So let us consider $x, x' \ne 0$. W.l.o.g. we may assume that $t \le t'$. If we had $t < t'$, then  $\lVert x' \rVert = \frac{1+t}{1+t'}\lVert x \rVert < \lVert x \rVert$. But then $2\lVert x' \rVert < 2\lVert x \rVert$ and $2-t' < 2-t$, thus $\alpha(x,t) < \alpha(x',t')$ which is impossible. We conclude $t = t'$ and $x = x'$.

*$H$ is surjective: Let $(y,s) \in D^n \times [0,1]$. (a) $\lVert y \rVert \le \frac{1+s}{2}$. Set $x = \frac{2-s}{1+s}y$ and $t = s$. Then $\lVert x \rVert =   \frac{2-s}{1+s}\lVert y \rVert \le  \frac{2-s}{1+s} \frac{1+s}{2} = \frac{2-s}{2} \le 1$ and $2\lVert x \rVert \le 2-s = 2-t$. Clearly $H(x,t) = (y,s)$. (b) $\lVert y \rVert \ge \frac{1+s}{2}$. Set $x = \frac{2-s}{2\lVert y \rVert}y$ and $t = 2\lVert y \rVert- 1 $. Then $\lVert x \rVert = \frac{2-s}{2\lVert y \rVert}\lVert y \rVert = \frac{2-s}{2} \le 1$ and $1  \ge t \ge s$. Hence $2\lVert x \rVert = 2-s \ge 2 - t$. Clearly $H(x,t) = (y,s)$.

*For $x \in S^{n-1}$ we have $H(x,t) = \left( \frac{1+t}{2}x , 0 \right)$, thus $H(S^{n-1} \times I) = \{ x \in \mathbb R^n \mid \frac{1}{2} \le \lVert x \rVert \le 1 \} \times \{0\}$. For $x \in D^n$ and $t = 0$ we have $H(x,t) = \left( \frac{1}{2}x , 0 \right)$, thus $H(D^n \times \{0\}) = \{ x \in \mathbb R^n \mid  \lVert x \rVert \le \frac{1}{2} \} \times \{0\}$. We conclude that $H$ is the desired homeomorphism.
