Does a circle retract to an open arc?

Question

I found the following question on Lima's Fundamental groups and covering spaces:

Show that a proper subset of $S^1$ is a retract of $S^1$ if and only if it is a circular arc.

I see how a proper closed arc, i.e. the image of a proper closed interval $[a,b]\subsetneq [0,1]$ under the quotient map $f(t) = e^{2\pi i t}$, is a retract of the circle.

Is this what the question meant? Or it possible to show that something like $S_1\setminus \{(1,0)\}$ also a retract of the circle? Because I'm confused as to how it is possible to have a continuous function $r : f([0,1]) \to f((0,1))$ that fixes $f((0,1))$.

edit: To clarify the notation, by $S^1$ I mean the set $\{(x,y)\in \mathbb{R}^2\ :\ x^2 + y^2 = 1\}$ with the subspace topology. I also forgot to add the curly brackets on $S_1\setminus \{(1,0)\}$ before. By that meant the circle without its rightmost point.

Answer 1

A retract of a Hausdorff space is closed (see this). And so in this context "circular arc" can only mean closed connected subset of $S^1$. Or equivalently (in case of $S^1$ only of course) an image of $[0,1]$ under some continuous map.

Which is slightly more than what you observe with $f([a,b])$ for $f(t)=exp(2\pi i t)$, because this does not cover arcs of the form $f([0,a]\cup [b,1])$. Unless you consider $f$ with say $\mathbb{R}$ as domain. Either way with this small change it indeed is the same.

Answer 2

Unfortunately Lima does not give a definition of an "arc of circle" (which you denote as a circular arc).

Usually an arc in a space $X$ is understood as a homeomorphic copy of $[0,1]$ in $X$, and certainly this is what Lima means. In other words, an arc in $S^1$ is the image of an embedding $e : [0,1] \to S^1$, or more generally the image of an embedding $e : [a,b] \to S^1$. The claim is

A proper subset $A \subset S^1$ is a retract of $S^1$ if and only if it is an arc in $S^1$.

1. Let $A$ be a retract of $S^1$. Then it is a compact connected subset of $S^1$. Pick $s_0 \notin A$ and $t_0 \in \mathbb R$ such that $s_0 =e^{2\pi i t_0}$. The map $p_0 : (t_0, t_0+2\pi) \to S^1 \setminus \{s_0\}, p_0(t) = e^{2\pi it}$, is a homeomorpism. Thus $p_0^{-1}(A)$ is a compact connected subset of $(t_0, t_0+2\pi)$ and therefore a closed subinterval $[a,b] \subset (t_0, t_0+2\pi)$. Therefore $A = p_0([a,b])$ is an arc in $S^1$.

2. Let $A$ be an arc in $S^1$. Pick $s_0 \notin A$ and $t_0 \in \mathbb R$ such that $s_0 =e^{2\pi i t_0}$. The map $p_0 : (t_0, t_0+2\pi) \to S^1 \setminus \{s_0\}, p_0(t) = e^{2\pi it}$, is a homeomorpism. Thus $p_0^{-1}(A)$ is an arc in $(t_0, t_0+2\pi)$ which means that there is an embedding $e : [0,1] \to (t_0, t_0+2\pi)$ such that $e([0,1]) = [a,b] = p_0^{-1}(A)$ for suitable $a, b$. Define $$r : [t_0, t_0+2\pi] \to [a,b], r(t) = \begin{cases} a + (\frac{a+b}{2}-a)\frac{t-t_0}{a-t_0} & t \le a \\ t \in [a,b] \\ b + (\frac{a+b}{2}-b)\frac{t-b}{t_0 + 2\pi -b } & t \ge b \end{cases}$$ This is a retraction such that $r(t_0) = \frac{a+b}{2} = r(t_0 + 2\pi)$. Thus it induces a retraction $\rho : S^1 \to p_0([a,b]) = A$: $\require{AMScd}$ \begin{CD} [t_0, t_0+2\pi] @>{r}>> [a,b] \\ @V{e^{2\pi it}}VV @VV{p_0}V \\ S^1 @>>{\rho}> A \end{CD}