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Munkres topology section 35(Tietze Extension Theorem) exercise 4-(c). The question is

Can you conjecture whether or not $S^1$ is a retract of $\mathbb{R}^2$?

I've read the answers in Is the unit circle $S^1$ a retract of $\mathbb{R}^2$?, but I don't know about algebra(algebraic topology) or Brouwer Fixed Point Theorem. Is there any way of proving this using at most calculus, analysis, elementary complex analysis and elementary topology(for example, things that appear in the first half of Munkres topology)? I already know from the exercise that $S^1$ is a retract of $\mathbb{R}^2 - \{0\}$.

Sphere
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2 Answers2

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The result that no such retract exists is equivalent to Brouwer's fixed point theorem (BFPT) for $n=2$. So, if you want to prove it without any algebraic topology, you have to look at proofs of BFPT which do not use algebraic topology. These do exist. There is one using Sperner's lemma on colouring triangle vertices (see Proofs from the Book) and a really nice one using the fact that the game of Hex can never be a draw (see article by David Gale). There are calculus ones, but I'm not familiar with them.

Teddy38
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A retract of a contractible space is contractible.

But $S^1$ is not contractible.

If you don't know about the fundamental group, or Brouwer's fixed point theorem, I still think there's an equivalent formulation in terms of extensions of maps from $S^1$ to the disc. It's equivalent to Brouwer, but says there's no extension of the identity function of $S^1$.

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    Yes, there is an equivalent formulation. See here https://math.stackexchange.com/questions/3843615/simply-connected-topological-space-a-detail-in-the-definition/3843635#3843635 – Sumanta Sep 30 '20 at 09:19
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    Also, retract of a contractible space is contractible https://math.stackexchange.com/questions/493723/a-retract-of-x-and-x-contractible-implies-a-contractible – Sumanta Sep 30 '20 at 09:21