I want to show that $\mathbb{R}^2$ and the open unit disk are homeomophic. This seems obvious as I can easily stretch the disk into the real plane, but I cannot think of a way to analytically do it. Does $$f(x,y)=(f_1(x,y),f_2(x,y)) =(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}). $$ get the job done?
[Math] homeomorphism from $\mathbb{R}^2 $ to open unit disk.
general-topologyreal-analysis
Related Solutions
Injectivity
$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means
$$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}= \tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|} $$
So we can assume $x\ne0$ and $y\ne0$. Then, taking norms, $$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr) $$ because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get $$ \frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2} $$ so $\|x\|=\|y\|$ and, finally $x=y$.
It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.
Surjectivity
$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have $$ z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|} $$ so $$ \|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr) $$ hence $$ \frac{\pi\|x\|}{2}=\arctan\|z\| $$ that is, $$ \|x\|=\frac{2\arctan\|z\|}{\pi}. $$ Thus the candidate is $$ x=\frac{2\arctan\|z\|}{\pi\|z\|}z $$ Verify it's the right one.
Continuity
The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well outside $0$
Are $F$ and its inverse continuous at $0$?
A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now $$ \|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\| =\tan\biggl(\frac{\pi\|x\|}{2}\biggr) $$ that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$.
Solve it directly. That is, solve $w = {z \over 1+ \|z\|}$ for $z$.
We see that $w=0$ iff $z=0$.
It is clear that $w$ and $z$ lie on the same line, so we can look for $z$ of the form $tw$ such that $w = {tw \over 1+ \|tw\|}$. It should be clear that if $w \neq 0$ then $t \ge 0$.
Then we have $1+t \|w\| = t$ and so $t={1 \over 1-\|w\|}$, from which we get $g(w) = {w \over 1-\|w\|}$.
We have $g(0) = 0$, $g$ is continuous, and we have $g(f(z)) = z$. It follows that $f$ is bijective.
Best Answer
You proposed homeomorphism doesn't quite work because it is not continuous at 0. Try instead proving that $$f (x)=\frac {x}{\Vert x\Vert+1} $$ does the job.