Here is my idea:

Suppose $f : \mathbb S^1 \to \mathbb S^1$ is continuous and injective.

If $f$ is not surjective, its image is homeomorphic to a subset of $\mathbb S^1 \setminus \{ c\} \cong \mathbb R$ (by considering the projection).

Then, the restriction of $f$'s codomain $g : \mathbb S^1 \to f(\mathbb S^1)$ is injective and surjective, so it is bijective.

Also, since $\mathbb S^1$ is compact and connected, its image under the continuous map is also compact and connected, so $f(\mathbb S^1)$ is homeomorphic to some closed interval $[a, b]$, which is simply connected.

Noting $f(\mathbb S^1)$ is Hausdorff and $\mathbb S^1$ is compact, by the closed map lemma, it follows $g$ is a homeomorphism. This contradicts that $\mathbb S^1$ is not simply connected, since $[a, b]$ is simply connected. Thus, $f$ must be surjective.

Does it work?


This is basically the same argument, but you can be a little bit shorter by noticing that any non-surjective continuous map $f:X \rightarrow \mathbb{S}^1$, where $X$ is a metric space, can be lifted to a map $\tilde{f} : X \rightarrow \mathbb{R}$. Of course, if $X=\mathbb{S}^1$, then $\tilde{f}$ cannot be injective, so $f$ is not injective.

  • $\begingroup$ Thanks--I don't see why $\tilde f$ or $f$ cannot be injective immediately, though. Also, does my idea work? $\endgroup$ Nov 18 '20 at 21:02
  • 1
    $\begingroup$ @TrashFailure it is the same idea as yours (how do you think he lifts the non-surjective map). You, however, work out the details more rigorously. The answer here relies on knowing that $S^1$ is not contractible. Yours can be used (with a little work) to prove the $S^1$ is not contractible. $\endgroup$
    – Tyrone
    Nov 18 '20 at 21:14

Your argument is completely correct, you should maybe add an argument why any compact connected subset of the reals is a closed interval (which is not difficult). Btw I don't see why the other proof presented here as an answer is a "little bit shorter", it is exactly the same argument and just explains less


May assume that our map $f$ takes $1$ to $1$. Consider the map $p\colon \mathbb{R}\to \mathbb{S}^1$, $t \mapsto \exp 2\pi i t $. There exists a "covering" $F\colon \mathbb{R} \to \mathbb{R}$ of $f$, such that $p\circ F= f \circ p$, and $F(0) = 0$. Now, since $p(F(t+1)) = p(F(t))$ for all $t\in \mathbb{R}$, we have $$F(t+1) - F(t) \equiv k \ \textrm {(an integral constant)}$$

In particular, $F(1) =k$. ( $k$ is the index of $f$).

If $k\ne 0$, then the image of $F$ contains the interval $[0,k]$ ( or $[k,0]$), and $p$ applied to this interval is the whole $\mathbb{S}^1$. Using $f\circ p = p\circ F$, we conclude $f$ is surjective.

If $k=0$, then $F(0)=F(1)=0$. Then it is easy to see that there exist $0<t_1<t_2<1$ such that $F(t_1) = F(t_2)$. It follows that $f(p(t_1)) = f(p(t_2)$, and so $f$ is not injective.


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