Homeomorphism between subset of 3-Sphere and Cartesian Product of unit disc and unit circle I am stuck trying to solve the following exercise.  
Let f: $S^1 \times S^1 \to D^2 \times S^1$ be defined by $f(x,y)=(y,x)$. 
Let
\begin{align*}
X= & \; \{(x_1,...,x_4) \in S^3 \mid x_1^2+x_2^2\geq x_3^2+x_4^2\}~\text{and}\\
Y= & \; \{(x_1,...,x_4) \in S^3 \mid x_1^2+x_2^2\leq x_3^2+x_4^2\}.
\end{align*}

a) Show that $X\cap Y$ is homeomorphic to $S^1\times S^1$ and that X and Y are homeomorphic to $D^2 \times S^1$.
b) Show that $(S^1 \times D^2)\cup_f(D^2 \times S^1)$ is homeomorphic to $S^3$. 

I was able to show the first part of a) but I'm stuck with the second part. Obviously I need a homeomorphism $h: X\to D^2\times S^1$. Since $1\geq x_1^2+x_2^2\geq x_3^2+x_4^2$ I can define $h(x_1,...,x_4)=((x_1,x_2),.)$ but I have problems defining the second component in such a manner that h is a homeomorphism. I probably have to use the fact that $x_1^2+x_2^2+x_3^2+x_4^2=1$, but I do not see how.
 A: Since $x_3^2+x_4^2 \leq x_1^2 + x_2^2 \leq 1$, it follows that $x_3^2+x_4^2\leq \frac{1}{2}$. Hence the projection $\pi_{34}:X\to \mathbb{R}^2$ defined by $\pi_{34}(x_1,x_2,x_3,x_4)=(x_3,x_4)$ has image 
$$\{(x_3,x_4)\in\mathbb{R}^2~|~0\leq x_3^2+x_4^2\leq \frac{1}{2}\},$$
which is a disk of radius $\frac{1}{2}$. Therefore the map $h_1:X\to D^2$ defined by $h_1(x_1,x_2,x_3,x_4) = (\sqrt{2}x_3, \sqrt{2}x_4)$ is a continuous surjection from $X$ onto the unit disk $D^2$. 
Again since $x_3^2+x_4^2 \leq x_1^2 + x_2^2$ and $x_1^2+x_2^2+x_3^2+x_4^2=1$, it follows that $0 < x_1^2+x_2^2$. Therefore the map $h_2:X\to S^1$ defined by $$h_2(x_1,x_2,x_3,x_4) = \left(\frac{x_1}{\sqrt{x_1^2+x_2^2}},\frac{x_2}{\sqrt{x_1^2+x_2^2}}\right)$$
always has positive denominator, and hence is a continuous surjection to the unit circle $S^1\subset \Bbb{R}^2$. 
Define the map $h:X\to D^2\times S_1$ by $h=(h_1,h_2)$, that is,
$$h(x_1,x_2,x_3,x_4) = \left((\sqrt{2}x_3,\sqrt{2}x_4),\left(\frac{x_1}{\sqrt{x_1^2+x_2^2}},\frac{x_2}{\sqrt{x_1^2+x_2^2}}\right)\right).$$
I will leave it to you to check that $h$ is a homeomorphism.
In part (b), we put together everything done previously. By definition, $S^3=X\cup Y$. You've already proven that $X\cap Y$ is $S^1\times S_1$. By the above paragraphs each of $X$ and $Y$ are homeomorphic to $D^2\times S^1$ and $S^1\times D^2$. Now just use the pasting lemma to extend the homeomorphisms $D^2\times S^1\to X$ and $S^1\times D^2\to Y$ to the entire space $D^2\times S^1\cup S^1\times D^2 \to S^3=X\cup Y$. 
