I am trying to prove that the cartesian product of the closed n dimensional disk with the closed m dimensional disk is homeomorphic to the n+m dimensional disk via a function that maps the boundary to the boundary homeomorphically.
Symbolically I am looking for a continuous function $f:D^n \times D^m \rightarrow D^{n+m}$ such that $f$ restricted to $S^{n-1} \times D^m \cup D^n \times S^{m-1} \stackrel{f}{\longrightarrow} S^{n+m-1}$ is also a homeomorphism. Here is what I have cooked up so far
First, define $k:(D^n \times D^m - \{0,0\}) \rightarrow (0,\infty)$ to be $k(x,y) = \frac{1}{min\{c_1,c_2\}}$ where $c_1|x| = 1$ and $c_2|y| = 1$ and if no such number exists set $c_i =\infty$. The purpose of this function is to generate the appropriate stretching factor for a point in $D^n \times D^m$ relative to the boundary. Now we can define $f(x,y) = (k(x,y)\frac{x}{|x,y|} , k(x,y)\frac{y}{|x,y|})$ for $(x,y) \ne (0,0)$ and $f(0,0)=(0,0)$.
I believe that this should work. However proving continuity is another issue, and there may be a more streamlined way to go about this. By the way, this problem comes from Strom's Modern Classical Homotopy Theory
I presume $D^n$ is the closed unit disc.
Define the closed cube by $$C^n=[-1,1]^n.$$ For each $n$ there is a homeomorphism $\Phi_n:C^n\to D^n$ (preserving boundaries). As $C^m\times C^n$ is $C^{m+n}$, then you can take your homeomorphism to be $\Phi^{m+n}\circ(\Phi_m^{-1}\times \Phi_n^{-1})$.
To construct $\Phi_n$, consider a nonzero point $x=(x_1,\ldots,x_n)\in C^n$. The set $D^n$ is $\{x\in\Bbb R^n:\|x\|_2\le1\}$, and $C^n$ is $\{x\in\Bbb R^n:\|x\|_\infty\le1\}$. Here $\|x\|_2=\sqrt{x_1^2 +\cdots+x_n^2}$ and $\|x\|_\infty=\max(|x_1|,\ldots,|x_n|)$. Define $\Phi_n(x)=(\|x\|_2/\|x\|_\infty)x$. Also define $\Phi_n(0)=0$. The continuity at $0$ is not entirely obvious: it follows from the boundedness of $\|x\|_2/\|x\|_\infty$ on the rest of the cube. Anyway, $\Phi_n$ is a continuous bijection between compact Hausdorff spaces, and so a homeomorophism.