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Is the following space, obtained by glueing a Cantor set worth of "hairs" to a closed disk in $\Bbb R^2$ contractible?

enter image description here

The obvious attempt of contracting the hairs to the root and then contracting the disk doesn't look continuous on the boundary of the disk.

Motivation: While looking for an example providing an answer to a question I posted a couple of days ago I stumbled on this construction of a compact contractible space which is nowhere locally connected, attributed to Robert Edwards but only published as an abstract of the AMS that I cannot find online. Showing that the space constructed by Edwards is contractible boils down to showing that the space depicted above is.

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    $\begingroup$ Seems like you can define a homotopy $f_t:X\to X$ with $t\in[0,2\pi]$ such that $f_0$ is the identity and $f_t$ rotates every point an angle of $t$ (when points of the hairs reach the "singular" point they keep rotating but in the boundary of the disk), and then $f_{2\pi}(X)$ is contained in the disk. $\endgroup$
    – Saúl RM
    Commented Apr 8, 2022 at 21:30
  • $\begingroup$ @SaúlRodríguezMartín Makes sense, you could post that comment as an answer! $\endgroup$
    – TopologicalDynamitard
    Commented Apr 9, 2022 at 8:29

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Calling the space $X$, you can consider the homotopy $f_t:X\to X$ such that $f_t$ rotates every point an angle of $-t$ around the origin. For points of the hairs you rotate them keeping them inside their hair, and after they reach the 'singular point', they keep rotating in the boundary of the disk.

Then $f_0$ is the identity and $f_{2\pi}(X)$ is contained in the disk, so $X$ is contractible.

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