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Are there some "elementary" ways to prove that circle can't be expressed as a two dimensioanl disks retract?

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    $\begingroup$ Welcome to MSE! Can you clarify why you might be interested in such a result? Currently this question looks low-effort, and is likely to be closed. However if you can add more context (as well as any ideas you have, or reasons you care) this will be more likely to attract upvotes, and possibly answers. $\endgroup$ Commented Sep 18, 2023 at 20:50
  • $\begingroup$ Let $f\colon D^{2}\to S^{1}$ be some such retract. Then the composition of $f$ with the antipodal inclusion of $S^{1}$ into $D^{2}$ manifestly lacks a fixed point, contradicting Brouwer's fixed point theorem, which in turn admits an elementary proof via Sperner's lemma. $\endgroup$
    – Rafi
    Commented Sep 18, 2023 at 20:52
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    $\begingroup$ Interesting question, but I have to say that it was a great relief to me to learn a little algebraic topology (which is useful anyway), rendering this an exercise. Ditto the Jordan Curve Theorem. Although I did always feel more engaged when driving a stick-shift, generally I do appreciate modern technology. :) $\endgroup$ Commented Sep 18, 2023 at 20:57
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    $\begingroup$ Yes, there is. The Brouwer fixed point theorem can be proven using real analysis only, Milnor did this in a nice write up on the Hairy Ball theorem $\endgroup$
    – FShrike
    Commented Sep 18, 2023 at 21:25
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    $\begingroup$ Keep in mind: analytic methods work great to rule out retractions that have some analytic properties, but to rule out continuous retractions you need something more. $\endgroup$
    – Lee Mosher
    Commented Sep 20, 2023 at 15:56

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