Is it true that any continuous injective map from a circle to circle is surjective?

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.

• Thanks--I don't see why $\tilde f$ or $f$ cannot be injective immediately, though. Also, does my idea work? Nov 18 '20 at 21:02
• @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. 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 $$0 such that $$F(t_1) = F(t_2)$$. It follows that $$f(p(t_1)) = f(p(t_2)$$, and so $$f$$ is not injective.