Show that the open unit disk in $\mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$ I have already perused this answer to a similar question but I still am unsure how exactly they went about proving it is a bijection. Would someone be able to elaborate? Also, how does one come up with the function $\frac{x}{1+||x||}$ to begin with? Thanks.
 A: $\dfrac x{1+|x|}$ Just scales $x$ by the factor $\dfrac 1{1+|x|}$.  Thus visually we are just moving along rays from the origin.
It is clear that it is injective, because if $x\ne y$, then we are on different rays, or $y=tx\implies \dfrac{y}{1+|y|}=\dfrac{tx}{1+t|x|}=\dfrac{x}{1/t+|x|}\ne\dfrac{x}{1+|x|}$, since $t\ne1$.
Surjectivity follows from the fact that every element of $\Bbb R^2$ is on some ray from the origin.  Just choose such a ray, and then adjust for the distance from the origin.  That is, given $y$,  $y=\dfrac{ty/|y|}{1+t}\implies (1+t)=t/|y|\implies  t=-|y|/(|y|-1)$.  So
$y=f(-y/(|y|-1))$.
A: It’s actually easier to see how to come up with the inverse of that bijection. The obvious approach to finding a bijection from the open disk to $\Bbb R^2$ is simply to expand the disk so that each radial line expands to an infinite ray in the same direction. This clearly requires that the closer we get to the unit circle, the more we have to expand. Since the distance from $x$ to the unit circle is $1-\|x\|$, the most straightforward way to do this is to multiply each vector $x$ by $\frac1{1-\|x\|}$: as $\|x\|$ increases from $0$ towards $1$, $\frac1{1-\|x\|}$ starts at $1$ and increases without bound. This gives us the map
$$h(x)=\frac{x}{1-\|x\|}\,.$$
Verifying that it’s a bijection is basically just a matter of showing that it has an inverse defined on $\Bbb R^2$. Suppose that $y\in\Bbb R^2$, and we want to find an $x$ such that $y=h(x)=\frac{x}{1-\|x\|}$. Clearly $x$ and $y$ are scalar multiples of each other, so there is an $\alpha\in\Bbb R$ such that $x=\alpha y$. Then $y=\frac{\alpha y}{1-\|\alpha y\|}$, so $\alpha y=\big(1-\|\alpha y\|\big)y$, and $\alpha=1-\|\alpha y\|$. And $\alpha$ is clearly positive, so $\alpha=1-\alpha\|y\|$, $\alpha\big(1+\|y\|\big)=1$, and $\alpha=\frac1{1+\|y\|}$. This is defined for every $y\in\Bbb R^2$, so $h$ is a bijection whose inverse is the map
$$f(x)=\frac{x}{1+\|x\|}\,.$$
A: HINT: Try to find the inverse. I guess the idea was to generalize $f(t)=\frac{t}{1+|t|}$ for higher dimensions.
