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I’m trying to prove that the open ball of radius 1 centered at the origin in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$. I believe the “shrinking map” from $\mathbb{R}^n$ to the ball given by $x \mapsto \dfrac{x}{1 + |x|}$ does the job, but I’m having trouble showing it’s a homeomorphism, particularly the “continuous inverse” part. What’s a good way to do this?

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To figure out the inverse, note that if $|x|=\lambda$, then

$$\left|\frac{x}{1+|x|}\right| = \frac{\lambda}{1+\lambda}.$$

Thus, given $y\in\mathbb{R}^n$ with $0\leq |y|\lt 1$, you want to find $\lambda$ such that $\lambda = (1+\lambda)|y|$. Letting $|y|=\mu$, we have

$(1-\mu)\lambda = \mu$, or $\lambda = \frac{1}{1-\mu}$.

So the map you want for the inverse is

$$y\longmapsto \frac{y}{1-|y|}.$$

Note that this is well-defined, since $0\leq |y|\lt 1$, so $0\lt 1-|y|\leq 1$. Also, the compositions are the identity:

$$\begin{align*}

x &\longmapsto \frac{x}{1+|x|}\\

&\longmapsto \left(\frac{1}{1- \frac{|x|}{1+|x|}}\right)\frac{x}{1+|x|}

= \left(\frac{1+|x|}{1+|x|-|x|}\right)\frac{x}{1+|x|}\\

&= \vphantom{\frac{1}{x}}x.\\

y &\longmapsto \frac{y}{1-|y|}\\

&\longmapsto \left(\frac{1}{1 + \frac{|y|}{1-|y|}}\right)\frac{y}{1-|y|}

= \left(\frac{1-|y|}{1-|y|+|y|}\right)\frac{y}{1-|y|}\\

&= \vphantom{\frac{1}{y}}y.

\end{align*}$$

Now simply verify that both maps are continuous.

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