This question started bothering me after working on an exercise. I know that there cannot be a contiuous bijection $S^1 \to (a,b)$ because if there was it would be a homeomorphism but $S^1$ and $(a,b)$ are not homeomorphic.

But the theorem that implies this is that a continuous bijection from a compact into a Hausdorff space is a homeomorphism. Hence it cannot be applied to the opposite direction.

I still suspect that the answer will turn out to be negative, i.e. there is no continuous bijection from $(a,b) \to S^1$ but I don't see how to prove it because the inverse is not required to be continuous. So somehow there still remains a faint possibility for such a map.

Please could someone help me resolve my confusion and tell me whether there is or is not a continuous bijection $(a,b) \to S^1$?


The other answer is more work than necessary. No 'big theorems' like invariance of domain are necessary in the 1D case (though if you look at the proof of invariance of domain, you'll see it's straightforward in dimension 1). I'm going to write absolutely none of the details because I'm absolutely lazy.

Let $f: (a,b) \to S^1 = \Bbb R/\Bbb Z$ be a continuous bijection. Consider $\lim_{x \to b} f(x)$. This exists by the injectivity hypothesis (can you prove this?) Similarly, so does $\lim_{x \to a} f(x)$. Hence there's an extension to $\tilde f: [a,b] \to S^1$. Because the previous map was surjective, $\tilde f(b)=f(c)$ for some $c \in (a,b)$. This is nonsense (again, why?)

In general any continuous bijection from one 1-manifold without boundary to another 1-manifold without boundary is a homeomorphism. This is also straightforward to prove without invariance of domain.

  • $\begingroup$ Thank you. Very very nice! I don't know what you mean by "by the injectivity hypothesis" -- doesn't it follow because continuous maps map Cauchy sequences to Cauchy sequences and in a compact space every Cauchy sequence has a limit in that space? $\endgroup$ Apr 7 '16 at 4:38
  • $\begingroup$ @self-learner: That's also a good proof. You can prove it without invoking compactness, though. $\endgroup$
    – user98602
    Apr 7 '16 at 4:40
  • 1
    $\begingroup$ What's that about continuous maps mapping Cauchy sequences to Cauchy sequences? What about the map $f:(0,1)\to[-1,1]$ defined by $f(x)=\cos\frac1x$ and the Cauchy sequence $\{\frac1{n\pi}\}$? $\endgroup$
    – bof
    Apr 7 '16 at 5:14
  • 2
    $\begingroup$ @self-learner The expression "maps nearby points to nearby points" suffers from vagueness. If you make it precise (using variables and "quantifiers" like "for all" and "there exists", and paying attention to the order of the quantifiers) then, depending on just how you do it, you end up with the notion of a continuous function or a uniformly continuous function. Uniformly continuous functions map Cauchy sequences to Cauchy sequences; just plain continuous functions don't necessarily. $\endgroup$
    – bof
    Apr 8 '16 at 4:49
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    $\begingroup$ I am not sure why $\lim_{x\rightarrow b} f(x)$ exists by the injectivity hypothesis, without invoking compactness. I mean, the map identity map from $(0,1)$ to itself is a continuous bijection, but $\lim_{x\rightarrow 1} f(x)$ doesn't exist. I realize you don't want to spoil it for self-learner, but if you don't mind spoiling it to me you can write it as a comment to some other post I have on MSE ;-) $\endgroup$ Sep 11 '18 at 14:39

No, there is no continuous bijection from $(a,b)$ to $S_1$. To show this, consider any bijective continuous map $f:(a,b)\rightarrow S_1$. Choose any $x\in S_1$. Note that $f$ is a bijective continuous map $(a,b)\setminus \{f^{-1}(x)\}\rightarrow S_1\setminus \{x\}$. However, $(a,b)\setminus \{f^{-1}(x)\}$ is homeomorphic to the disjoint union of two open intervals and $S_1\setminus \{x\}$ is homeomorphic to a single open interval.

Thus, if such an $f$ existed, we would have that some continuous bijective $g$ mapped $(0,1)\cup (1,2)$ to $(0,1)$. However, this is impossible, since invariance of domain (which is very easy to prove in one dimension) would imply that such a $g$ was an open map and thus a homeomorphism, but clearly $(0,1)\cup (1,2)$ and $(0,1)$ are not homeomorphic.

Another way, which captures somewhat different intuition, but uses more powerful tools, would be to take any continuous $f:(a,b)\rightarrow S_1$ and then consider the map $g:(a,b)\times (0,\infty)\rightarrow \mathbb R^2$ defined by $g(x,y)=yf(x)$ where we regard $S_1$ as a subset of the plane. Then, if $f$ is surjective, the image of $g$ is $\mathbb R^2\setminus \{0\}$. However, $(a,b)\times (0,\infty)$ is not homeomorphic to $\mathbb R^2\setminus \{0\}$, so $g$ cannot be a homomorphism and thus, by invariance of domain, may not be injective. However, $g$ would be injective if $f$ was, implying that $f$ is not injective either.

  • $\begingroup$ I do not understand why this was downvoted. $\endgroup$
    – user98602
    Apr 7 '16 at 4:36
  • $\begingroup$ I did not downvote any of the answers here. I am very thankful that everyone here is trying to help me! $\endgroup$ Apr 8 '16 at 4:41

Suppose $f:(0,1)\to S^1$ is such a function.

Let $y\in S^1$. Let $x=f^{-1}(y)$.

Then $(0,1)\setminus \{x\}=(0,x)\cup (x,1)$.

Now $f[(0,x)]$ and $f[(x,1)]$ are nontrivial disjoint connected sets that union to $S^1\setminus \{y\}=(0,1)$.

So WLOG $f[(0,x)]=(0,z)$ and $f[(x,1)]=[z,1)$ for some $z\in (0,1)$.

There exists $p\in (x,1)$ such that $f(p)=z$.

Then $(x,p]$ and $[p,1)$ map to nondegenerate intervals $A$ and $B$ in $[z,1)$ both containing $z$. So there exists $z'\in (A\cap B)\setminus \{z\}$. But then $f$ must map two different points to $z'$. Contradiction.


Recall (from high school?) that any two distinct points $P$ and $Q$ of $S^1$ define two arcs of $S^1$ with these two points serving as the endpoints of both of these arcs. We can talk about both closed and open arcs depending on whether we include or exclude these endpoint on any arc. Also, if we remove any one point from $S^1$ we have an open arc.

This can be related to the theory/machinery of topological spaces:

Proposition 1: Let $f: [a,b] \to S^1$ be a continuous and injective function. Then the image of $f$ is an arc of $S^1$ that is also a closed set. Moreover, the restriction of $f$ to the open interval $(a,b)$ is an open set in $S^1$.
Proof: Exercise.

Proposition 2: Let $f: (a,b) \to S^1$ be a continuous and injective function. Then $f$ can't be a surjection. Moreover, the image of $f$ is an open arc.
Proof Sketch:
Consider the increasing chain of of open intervals

$\tag 1 U_n = \left(\frac{(n-1)a + b}{n}, \frac{a + (n-1)b}{n}\right) \subset (a,b) \; \text{ for } n \ge 3$

The union of the $U_n$ is equal to entire domain $(a, b)$ of $f$ and the images $f(U_n)$ form an open cover of the image of $f$.

Assume that $f$ is a surjective mapping. We have an an open covering $f(U_n)$ of the image $f\left((a,b)\right) = S^1$. Since $S^1$ is compact we can find a finite subcover. But no element in the increasing chain $f(U_n)$ can ever be equal to the image of $f$, giving us an absurdity.


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