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In the topological sense, I understand that the unit circle $S^1$ is a retract of $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ where $\mathbb{0}$ is the origin. This is because a continuous map defined by $r(x)= x/|x|$ is a retraction of the punctured plane $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ onto the unit circle $S^1 \subset \mathbb{R}^2 \backslash \{\mathbb{0}\}$. Does this mean that $S^1$ is not a retract of $\mathbb{R}^2$? I would appreciate some clarification here.

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It is worthy to mention that all answers given below can be easily generalised to higher dimensions, ie. there is no retraction of $r:\mathbb{R}^n\to S^{n-1}$ for $n\ge2$. The proofs are identical. Also there is no retraction of $\mathbb{R}$ onto $S^0$, since the latter is disconnected and we consider only continuous retractions. – Damian Sobota Nov 21 '11 at 20:05
up vote 10 down vote accepted

No, you cannot conclude that $S^1$ is not a retract of $\mathbb R^2$ that way. To prove that something is not a retract usually requires more machinery, and algebraic topology is more or less designed to be helpful for this. I'll explain an argument using the fundamental group $\pi_1$, but one could use other functors for the same purpose (homology being the most obvious alternative)

If a subspace $Y\subseteq X$ is a retract of $X$, there is a retraction $r:X\to Y$ such that the composition $r\circ i$ with the inclusion $i:Y\to X$ is the identity of $Y$. If we pick a base point $x_0\in Y$, then the inclusion map $i$ induces an homomorphism $\pi_1(i):\pi_1(Y,x_0)\to\pi_1(X,x_0)$ such that $$\pi_1(r)\circ\pi_1(i)=\pi_1(r\circ i)=\pi_1(\mathrm{id}_Y)=\mathrm{id}_{\pi_1(Y)}.$$ In particular, the map $\pi_1(i)$ is injective.

But for any choice of $x_0\in S^1$, the map $\pi_1(i):\pi_1(S^1,x_0)\to\pi_1(\mathbb R^2,x_0)$ is not injective. Indeed, $\pi_1(S^1,x_0)$ is a non-trivial group while $\pi_1(\mathbb R^2,x_0)$ is trivial.

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Indeed there is no retraction $r: \mathbb{R^2} \rightarrow S^1$ because if $\iota :S^1 \rightarrow \mathbb{R^2}$ is the inclusion you would have a monomorphism $\iota^* : \pi(S^1) \rightarrow \pi(\mathbb{R^2})$ between the fundamental groups, i.e. a monomorphism $\mathbb{Z} \rightarrow \{0\}$ which would be absurd.

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Should be $\mathbb R^2$. – Rasmus Nov 21 '11 at 18:53

You can use the Brouwer Fixed Point Theorem to show that $S^1$ is not a retraction of the unit disk, and hence not a retraction of the entire plane (since, if $X\subset Y\subset Z$, and $j:Z\rightarrow X$ is a retraction, then the restriction $j_{|Y}:Y\rightarrow X$ is a retraction.)

Assume there is a retraction $f:D^2\rightarrow S^1$. Define $g(x):D^2\rightarrow D^2$ as $g(x)=-f(x)$. Then if $x\in S^1$, $f(x)=x$, by the condition that $f$ is a retraction, so $g(x)=-x$, and hence $g(x)\neq x$. If $x\notin S^1$, then $g(x)\in S^1$, and so $g(x)\neq x$. So there are no fixed points for $g$, contradicting Brouwer.

Now, usually, Brouwer is proven the opposite way - using Algebraic Topology to show that there can be no retraction of $D^2\rightarrow S^1$, and then showing that if $g:D^2\rightarrow D^2$ has no fixed point, then you can get a retraction of $D^2$ to $S^1$.

However, Brouwer has other, non-Algebraic proofs. Even constructive proofs.

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