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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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    @ZhenLin Then the question would be how does $D^n\cong I^n$ ?2012-01-21

1 Answers 1

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Put the unit disk inside the cube $[-1,1] \times \ldots \times [-1,1]$ (with $n$-factors). Then project radially.

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    Yeah, that's the intuition, I get it. But can you kindly provide an explicit function that is an explicit expression?2018-10-27