Continuous bijection which is not a homeomorphism.

The function $f$ is surjective becasue if $(x,y)\in S^1$, there is a $\theta\in[0,2\pi)$ such that $(x,y)=(\cos\theta,\sin\theta)$; just take $\theta=\arccos x$ if $y\geqslant0$ and $\theta=2\pi-\arccos x$ otherwise.

And $f^{-1}$ is discontinuous because $\lim_{n\to\infty}\left(\cos\left(2\pi-\frac1n\right),\sin\left(2\pi-\frac1n\right)\right)=(1,0)=f(0)$, but $\lim_{n\to\infty}f^{-1}\left(\cos\left(2\pi-\frac1n\right),\sin\left(2\pi-\frac1n\right)\right)$ doesn't exist (in $[0,2\pi)$).

To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_n\to (1,0)$ but $f^{-1}(y_n)\to 2\pi\ne f^{-1}(1,0)=0.$ Consider the sequence of points given by $$ (\cos (2\pi-n^{-1}), \sin (2\pi-n^{-1}) ) $$ for example.