$S^1$ has no deformation retracts other than itself

Question

I am trying to prove that if $A \subset S^1$ is a deformation retract of $S^1$, then $A=S^1$.

$A$ being a deformation retract of $S^1$ means that there exists a continuous map $r \colon S^1 \to A$ such that $r\circ i = id_A$ and $i \circ r \simeq id_{S^1}$, where $i$ is the inclusion. As $i \circ r \simeq id_{S^1}$, there exists a continuous map $H \colon S^1 \times I \to S^1$ such that $H((x,y),0)=(x,y)$ and $H((x,y),1)=i\circ r (x,y)$ for all $(x,y) \in S^1$. I am trying to prove that $i \circ r(x,y) =(x,y)$, or, equivalently, that $f=id_{S^1}$, where $$\begin{aligned}[t]f \colon S^1 &\longrightarrow S^1 \\ (x,y) &\longmapsto f(x,y)=H((x,y),1) \end{aligned}$$ This would imply $A=S^1$ (as $r \colon S^1 \to A$ would be a homeomorphism) but I have not been able to show that $f=id_{S^1}$, so it is possible that $f=id_{S^1}$ is not necessarily true in the first place. I would appreciate any hints on that.

Answer 1

Hint number $1$: consider that if $A\neq S^1$ then $i_\ast:\pi_1(A)\to\pi_1(S^1)$ factors as $\pi_1(A)\to\pi_1(S^1\setminus\{\zeta\})\to\pi_1(S^1)$.

Hint number $2$: use instead the classification of connected subsets of $\Bbb R$ to realise that all retracts of $S^1$ are subarcs or the whole circle itself. Can $S^1$ deform onto a subarc? You'll probably need knowledge of $\pi_1(S^1)$ or similar technology to answer that.

Answer 2

In the case of $A=\{*\}$ being a point the statement can be seen as: $S^1$ is not contractible. You won't be able to show that without some higher level machinery like fundamental group, or homology or Brouwer fixed point theorem or something like that. It is not a simple homework exercise.

On the other hand, if you already know that then the conclusion easily follows because every proper retract (or even proper connected subspace) of $S^1$ is an arc. So it is contractible. And deformation retraction is a special case of homotopy equivalence.