Showing that $D^2\setminus S^1$ is diffeomorphic to $\mathbb{R}^2$ I'm studying some manifolds and I'm stuck at the exercise 1.5c of Tu's Introduction To Manifolds.

There he asks us to verify that the function $h \colon D^2 \setminus S^1 \to \mathbb{R}^2$ given by $$h(a,b) = \left( \frac{a}{\sqrt{1-a^2-b^2}}, \frac{b}{\sqrt{1-a^2-b^2}}\right)$$ is a diffeomorphism.

How one should go about that? I know that we need to check that both $h$ and $h^{-1}$ are bijective smooth maps, but I can't think on how I should go about proving that all derivatives exist and are continuous, I can't find a general formula.
 A: Finding the partial derivatives of $h_1(a,b)=\frac{a}{\sqrt{1-a^2-b^2}}$ and $h_2(a,b)=\frac{b}{\sqrt{1-a^2-b^2}}$ is just an application of the quotient rule. Computing each:
$$\frac{\partial h_1}{\partial a}=\frac{-b^2+1}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_1}{\partial b}=\frac{ab}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_2}{\partial a}=\frac{ab}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_2}{\partial b}=\frac{-a^2+1}{(1-a^2-b^2)^{3/2}}$$
The numerators are all everywhere continuous and the denominators are only zero on $S^1$. Thus $h$ is differentiable on all of $D^2\setminus S^1$. Moreover, when we compute the determinate of the Jacobian we get:
$$\frac{\partial h_1}{\partial a}\frac{\partial h_2}{\partial b}-\frac{\partial h_1}{\partial b}\frac{\partial h_2}{\partial a}=\frac{a^2b^2-a^2-b^2+1-a^2b^2}{(1-a^2-b^2)^{3}}=\frac{1}{(1-a^2-b^2)^{2}}$$
From this we see that the determinate is non-singular on all of $D^2\setminus S^1$, and so by the inverse function theorem $h$ has an everywhere differentiable inverse $h^{-1}:h(D^2\setminus S^1)\to D^2\setminus S^1$. Thus, all that is left is to show that $h(D^2\setminus S^1)=\mathbb{R}^2$. Notice that the function $h(a,b)=\frac{1}{\sqrt{1-a^2-b^2}}(a,b)$ just scales the vector $(a,b)$ by $\frac{1}{\sqrt{1-r^2}}$ where $r$ is the magnitude of $(a,b)$. Picking any $\hat{s}\in S^1$ we can thus view $h$ as a map from $[0,1)\times S^1$ to $\mathbb{R}^2$ by $h(r,\hat{s})=\frac{r}{\sqrt{1-r^2}}\hat{s}$ or just as a map $h_\hat{s}$ from $[0,1)$ to $[0,\infty)$ via $h_\hat{s}(r)=\frac{r}{\sqrt{1-r^2}}$. Notice $h_\hat{s}(0)=0$ for all $\hat{s}$ so this way of viewing $h$ is well defined. $h_\hat{s}$ is a surjection and so we see that $h$ is a surjection from $D^2\setminus S^1\to\mathbb{R}^2$.
