# Is there a continuous, surjective map from the unit disc to the unit circle?

Is there a continuous, surjective, map from the unit disc with boundary, in two dimensions $$D_ 1 = \{ (x , y) \mid x^ 2 + y ^ 2 \leq 1 \}$$ to its boundary $$S_ 1 = \{ (x , y) \mid x^ 2 + y ^2 = 1 \}?$$

I am having trouble figuring out what to do with the origin point. My first guess was the projection below, but it's not continuous. $$r e ^ { i \theta } \mapsto e ^ { i \theta} , r \neq 0, 0 \mapsto 1$$

• What about $f(x,y)=(1,0)$? Commented Apr 4, 2022 at 14:11
• Apologies, I missed out the surjective condition in my question! Commented Apr 4, 2022 at 14:14
• why is the polar projection not continuous? You have a problem at zero since it maps everywhere? Commented Apr 4, 2022 at 14:17
• What about something like $f(x,y)=e^{\pi i x}$? Just a parametrization of the circle by the interval $[-1,1]$, you drop the $y$ value and ignore it completely? Maybe you're also missing a hypothesis in the question, that the map is constant on the boundary of the disk? Commented Apr 4, 2022 at 14:21
• Squish the disk into a line segment, and then wrap the line segment around the circle. Unless you wanted the edge of the disk to stay fixed, in which case there isn't one. Commented Apr 4, 2022 at 14:22

Notice that there is such a continuous surjection from the unit inerval, namely, $$f:[0,1]\to S^1$$ given by $$f(x)=e^{2\pi i x}$$. You can also find a continuous surjection $$g:D_1\to [0,1]$$, for instance, project onto the $$X$$-axis to get the interval $$[-1,1]$$, and then smash the left half of the interval onto the interval $$[0,1]$$.
Edit: as said in the comments, you can make a more direct description by defining $$f:[-1,1]\to S^1$$ by $$f(x)=e^{\pi i x}$$, so you don't need to factor by $$[0,1]$$.