An example of a continuous bijection $f$ where $f^{-1}$ isn't continuous Give an example of a continuous bijection $f:\mathbb{R}^2\rightarrow B=\left \{(x,y,z);(\sqrt{x^2+y^2}-1)^2+z^2=1\right \}$ such that $f^{-1}$
: B → $\mathbb{R}^2$
fails to be continuous.
I really don't know where to start here, I appreciate any help.
 A: Clearly, no continuous bijection mapping $\mathbb R^2 \to B$ can have a continuous inverse, since the two sets are not homeomorphic. Thus, the question is, can you find some continuous bijection mapping $\mathbb R^2 \to B$?
Here is a hint on how you might find one such thing.  The set $B$ looks something like so:

Note that double cusp at the origin of $\mathbb R^3$. You need a function $f:\mathbb R^2 \to \mathbb R^3$ that maps some point to $(0,0,0)$ and satisfies
$$\lim_{x,y\to\infty}f(x,y) = (0,0,0).$$
Disks centered at the the point that maps to $(0,0,0)$ and with increasing radius should spread out over your set $B$ something like so:


We can construct such a function explicitly as follows. Let's start with 
$$
f_1(r,\theta) = 2 \sin (r) \text{sgn}(\sin (r))
\left(
\begin{array}{ccc}
 \cos (\theta ) & -\sin (\theta ) & 0 \\
 \sin (\theta ) & \cos (\theta ) & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
 \sin (r) \\
 0 \\
 \cos (r) \\
\end{array}
\right).
$$
Note that
$$
f_1(r,0) = 2 \sin (r) \text{sgn}(\sin (r)) \;
\langle
 \sin (r),
 0,
 \cos (r)
\rangle
$$
for $-\pi < r < \pi$ parametrizes two circles in the $xz$-plane meeting at the origin like so:

Again, the two circles together are exactly the image under $f_1(\cdot,0)$ of the interval $(-\pi,\pi)$. The thick portion of the image above shows the image of $[-2.5,2.5]$ under $f_1$. The matrix in the definition of $f_1$ rotates about the $z$-axis through the angle $\theta$. Thus, $\,f_1$ maps an open disk of radius $\pi$ described in polar coordinates with $-\pi < r < \pi$ and $0<\theta\leq\pi$ onto $B$. Finally, we define 
$$f(r,\theta) = f_1(2\arctan(r),\theta)$$
to obtain a continuous bijection from the plane to $B$.
