Homeomorphism of the unit disk onto itself which does not extend to the boundary It is well known that any conformal mapping of the unit disk onto itself extends to the unit cirle. 
However, is there an homeomorphism of the unit disk onto itself which does not extend to a continuous function on the closed unit disk?
If yes, can you give an explicit one?
Thank you,
Malik

This following gives an example, probably:
Let $H=\{(x,y)\in\mathbb R^2:y>0\}$ be the upper half plane in $\mathbb R^2$ and let $\bar H$ be its closure. Let me give you an homeo $f:H\to H$ which does not extend to an homeo $\bar f:\bar H\to\bar H$, and which fixes the point of infinity on the $x$-axis. Then you can conjugate with an homeo $D\to H$ from the open unit disk which extends to the boundary (apart from one point...)
Let $\phi:\mathbb R\times(0,\infty)\to\mathbb R$ be given by $\phi(x,t)=\frac1{\sqrt{4\pi t}}\exp\left(-\frac{x^2}{4t}\right)$, and let $h:\mathbb R\to\mathbb R$ be a positive smooth function with support on $[-2,2]$ and such that it is constantly $1$ on $[-1,1]$. Define now $g:H\to H$ so that when $x\in\mathbb R$ and $y>0$ we have $$
g(x,y)=\int_{-\infty}^\infty\phi(x-\xi,y)(1-h(\xi))\,\mathrm d\xi.
$$ This is a solution of the one-dimensional head equation. Properties of that equation (which one can easily show in this case!) imply that $f$ extends continuously to a map $\bar g:\bar H\to\mathbb H$.
Now consider the map $f:(x,t)\in H\mapsto (xg(x,t),t)\in H$. This is an homeomorphism, and it extends to the map $\bar f:(x,t)\in\bar H\mapsto (x\bar g(x,t),t)\in\bar H$, which is not an homeo.
