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I've proven that "If $A$ is a retract of $X$ and $X$ is contractible, then $A$ is also contractible."

Can I use this to prove the circle $S^1$ isn't a retract of the disk $D^2$ or of the plane $\Bbb R^2$?If so, how can I do that?

My trial:

$(1)$ I was allowed to take this theorem:

The circle $S^1$ isn' t contractible,

for granted and I can use it. So I decided to prove that the disk $D^2$ (and the plane $\Bbb R^2$ )is contractible (which can be done because any convex space is homotopically equivalent to the space of the one point (by the straight line homotopy and because $D^2$ and $\mathbb{R^2}$ are convex spaces)) and to assume that $S^1$ is a retract of the disk $D^2$ ( or of the plane $\Bbb R^2)$ and then, by the problem I've proven, I should conclude $S^1$ is contractible, which contradicts the theorem above that I took for granted .... am I correct in this proof?

$(2)$ I have found a solution in this link which uses the idea of the fundamental group and Brouwer fixed point theorem, but, unfortunately, these tools I haven't studied yet.

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1 Answers1

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By contradiction is fine, in detail:

  1. Assume $S^1$ is a retract of $D^2$.
  2. $D^2$ is contractible (by straight line segments to the origin, as you say).
  3. Quoted theorem applied to $1$ and $2$: $S^1$ is contractible.
  4. Fact given to you: $S^1$ is not contractible.

Hence contradiction from $3$ and $4$ and so $1$ must be false: $S^1$ is not a retract of $D^2$.

Another proof could use the Brouwer fixed point theorem:

  1. Assume $S^1$ is a retract of $D^2$.
  2. Theorem by Brouwer: every $f: D^2 \to D^2$ that is continuous has a fixed point. (i.e. $D^2$ has the FPP (fixed point property)).
  3. Theorem: if $X$ has the FPP and $A$ is a retract of $X$, then $A$ has the FPP too. (If $g: A \to A$ is continuous, and $r$ is a retraction, apply the FPP to $g \circ r: X \to A \subseteq X$, etc.)
  4. $1$ to $3$ together imply $S^1$ has the FPP.
  5. Fact: $S^1$ does not have the FPP (consider the rotation around $90$ degrees either way, e.g.).
  6. Contradiction between $4$ and $5$ shows $1$ must be false again.
Henno Brandsma
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