Is the Interior of Hyperbola homeomorphic to $\mathbb{R^{2}}$?

Question

I know that the interior of the unit disk is homeomorphic to $\mathbb{R^{2}}$ by the mapping $(r,\theta)\to(\tan(\frac{r\pi}{2}),\theta)$. I am struggling to come up with a homeomorphic map from the interior of the hyperbola $x^{2}-y^{2}=1$ to $\mathbb{R}^{2}$ or to the interior of the unit disk. I think I am struggling mainly because I cannot find suitable coordinates to describe the interior. For example in the unit disk I just could write it as $r<1$.

By interior of the hyperbola I mean $\{(x,y):x^{2}-y^{2}<1\}$

Can anybody help me with a map from it to the disk or to $\mathbb{R}^{2}$?

I am very very new to homeomorphisms (Just about learnt the definitions).

Answer 1

Let $H$ be your region. If $(x,y)\in H$, then $x\in\left(-\sqrt{y^2+1},\sqrt{y^2+1}\right)$. And, given $a>0$, a homeomorphism between $(-a,a)$ and $\Bbb R$ is the one defined by$$\begin{array}{ccc}(-a,a)&\longrightarrow&\Bbb R\\x&\mapsto&\dfrac x{1-\left|\frac xa\right|}.\end{array}\tag1$$So, consider\begin{array}{ccc}H&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\left(\dfrac x{1-\left|\frac x{\sqrt{y^2+1}}\right|},y\right).\end{array}

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Another possibility is to consider$$\begin{array}{ccc}(-a,a)&\longrightarrow&\Bbb R\\x&\mapsto&\tan\left(\frac{\pi x}{2a}\right)\end{array}$$instead of $(1)$, which leads to\begin{array}{ccc}H&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\left(\tan\left(\dfrac{\pi x}{2\sqrt{y^2+1}}\right),y\right).\end{array}

Answer 2

Hint
For a conformal parameterization, study the well-known picture

made up of confocal hyperbolas and ellipses.