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Is there a continuous function $f:S^1 \to \mathbb R$ which is one-one?

Ayman Hourieh
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La Rias
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3 Answers3

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Suppose such a function $f$ exists. $f(S^1)$ must be connected and compact. $f$ is one-one so it cannot be constant. It follows that $f(S^1) = [a, b]$ for $a \ne b$.

Pick a point $s \in S^1$ such that $f(s) \not\in \{a, b\}$. $S^1 - \{s\}$ is connected, but $[a, b] - \{f(s)\}$ is not. We conclude that such a function $f$ cannot exist.

Alternatively, if you're familiar with the concept of the fundamental group: The restriction $f : S^1 \to f(S^1)$ must be a homeomorphism (being a continuous bijection from a compact space onto a Hausdorff space). But we know that $\pi_1(S^1) = \mathbb Z$ whereas any connected subset of $\mathbb R$ is contractible; hence it has a trivial fundamental group.

Ayman Hourieh
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Suppose such a function exists. Let $u : [0,1] \to S^1$ be a suitable path that traces around the circle, and consider $g = fu$. This $g : [0,1] \to \mathbb R$ is one-to-one except that $g(0) = g(1)$.

Consider $y = g(1/2)$. It must be either greater than or less than $g(0)$. Pick some value $z$ between $g(0)$ and $y$. Apply the intermediate value theorem on both the intervals $[0,1/2]$ and $[1/2,1]$ and show that there must be some $c$ and $d$ in $(0,1/2)$ and $(1/2,1)$ respectively with $g(c) = g(d) = z$, contradicting $g$ being one-to-one.

In simple terms: the function must go either up or down from where it starts, but then it needs to go back through territory it's already visited to get back to the first point.

Ben Millwood
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    Or you could argue that any one-to-one continuous function $f\colon [0,1]\to\mathbb R$ is monotone, using IVT. Then $f(0)\neq f(1)$ 'nuff said. – Cheerful Parsnip Nov 16 '13 at 16:42
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It cannot be 1-1 because of the one-dimensional version of Borsuk-Ulam: for every continuous $f: S^1 \rightarrow \mathbb{R}$ there exists an $x \in S^1$ such that $f(x) = f(-x)$.

The proof in the one-dimensional case (it's true for all $S^n$ and $\mathbb{R}^n$ but harder to prove) is not very hard: define $g(x) = f(x) - f(-x)$. Suppose for some $p \in S^1$, $g(p) \neq 0$ (otherwise we would be done anyway), then $g(-p) = -g(p)$ by definition and so then $g$ assumes positive and negative values in $\mathbb{R}$, and so also assumes the value $0$ by the intermediate value theorem (we use the connectedness of $S^1$). And we are done for such a value: $g(x) = 0$ iff $f(x) = f(-x)$.

Henno Brandsma
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