# $D^n\times D^m\cong D^{n+m}$?

Let $D^n=\{x\in \mathbb{R}^n : ||x||\le 1\}$ the unit closed ball in dimension $n$, and $I=D^1=[0,1]$.

We have $D^1\times D^1=I\times I\cong D^2$.

Is it true that

$$D^n\times D^m\cong D^{n+m}$$

for all $n,m\ge 0$ ? If so, where does the homeomorphism comes from ?

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Yes. Simply find a homeomorphism $D^n \cong I^n$ for each $n$, and then it's obvious... –  Zhen Lin Jan 21 '12 at 1:11
@ZhenLin Then the question would be how does $D^n\cong I^n$ ? –  Klaus Jan 21 '12 at 1:54
Put the unit disk inside the cube $[-1,1] \times \ldots \times [-1,1]$ (with $n$-factors). Then project radially.