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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. $$ r e ^ { i \theta } \mapsto e ^ { i \theta} , r \neq 0, 0 \mapsto 1 $$

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    $\begingroup$ What about $f(x,y)=(1,0)$? $\endgroup$ Commented Apr 4, 2022 at 14:11
  • $\begingroup$ Apologies, I missed out the surjective condition in my question! $\endgroup$ Commented Apr 4, 2022 at 14:14
  • $\begingroup$ why is the polar projection not continuous? You have a problem at zero since it maps everywhere? $\endgroup$
    – gt6989b
    Commented Apr 4, 2022 at 14:17
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    $\begingroup$ 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? $\endgroup$
    – Anthony
    Commented Apr 4, 2022 at 14:21
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    $\begingroup$ 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. $\endgroup$
    – JonathanZ
    Commented Apr 4, 2022 at 14:22

1 Answer 1

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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 $[0,1]$.

Overall, this is a composition of continuous surjections, so it is a continuous surjection.

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 $[0,1]$.

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  • $\begingroup$ Thanks! I'm also just curious if there's a map which preserves the boundary, such that the restriction on the boundary is just the identity map. $\endgroup$ Commented Apr 4, 2022 at 15:20
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    $\begingroup$ @AfiqHatta No there isn't. See proposition 5.3.2 on page 237 here mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/… for a short proof of the smooth case. This is Brouwer's no-retraction theorem. $\endgroup$
    – Mason
    Commented Apr 5, 2022 at 0:03
  • $\begingroup$ @AfiqHatta Mason has already answered your comment. Would you mind accepting my answer? $\endgroup$
    – Javi
    Commented Apr 5, 2022 at 21:18

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