# Is the unit circle $S^1$ a retract of $\mathbb{R}^2$?

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.

• 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. Commented Nov 21, 2011 at 20:05

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.

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.

• Should be $\mathbb R^2$. Commented Nov 21, 2011 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.

Intuitively, a (continuous) retraction of $R^2$ onto $S^1$ would not be possible, because there would have to be a "tear" in the disc to accomplish this. The tear wouldn't be continuous, as nearby points need to go to nearby points. .. Using the notion of topology as rubbersheet geometry, "tearing" (of the rubber) isn't allowed. .. if the plane was punctured, on the other hand, it could be deformed onto the circle, (continuously)...

• Of course it (S^1) is also not a retract of the unit disc (D^2).
– user403337
Commented Jul 13, 2017 at 4:54