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Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous, and locally 1-1. I want to show it is globally 1-1 (without assuming the existence of $f'$).

The intermediate value theorem implies that $f$ is locally strictly monotonic. Intuitively, I would like to show that if $f(a)=f(b)$, then somewhere between $a$ and $b$, $\,f$ must "switch directions", but I haven't had any traction with this strategy.

Any ideas?

Eric Auld
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    Have you already a theorem that says that a continuous function assumes a maximum on a closed interval $[a,, b]$? – Daniel Fischer Jul 15 '13 at 22:31
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    If $f(a)=f(b)$, look at the maximum $M$ of $f$ in $[a,b]$. (If it is $a$, then look at the minimum.) Now consider a neighborhood of $M$. – Andrés E. Caicedo Jul 15 '13 at 22:31
  • Thanks, guys, that solves it! – Eric Auld Jul 15 '13 at 22:34
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    @GitGud Presumably, it means every point in the domain has some neighborhood on which $f$ is one-to-one. – augurar Jul 15 '13 at 22:50
  • @DanielFischer It seems we have the same theorem for $f:\mathbb{R}^d \to \mathbb{R}$, or indeed if $f$ is defined on any convex subset of $\mathbb{R}^d$, since if $f(p)=f(q)$, then consider the line from $p$ to $q$...$f$ takes a maximum on that line, and $p_{\max}$ cannot have a neighborhood on which it is 1-1. Is this correct? – Eric Auld Jul 15 '13 at 23:05
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    @EricAuld For $d > 1$, a continuous function $f \colon U \to \mathbb{R}$, where $U \subset \mathbb{R}^d$ is open and non-empty can never be locally $1$-$1$. Consider a small closed path near any point. If you consider functions $f \colon U \to \mathbb{R}^d$, those can be locally $1$-$1$ without being globally $1$-$1$. (If you already know it, consider the complex exponential function, otherwise $(x,,y) \mapsto (e^x\cos y,, e^x\sin y)$.) – Daniel Fischer Jul 15 '13 at 23:12
  • Very helpful, thanks! – Eric Auld Jul 15 '13 at 23:14

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You could prove the similar statement that a continuous function $f: \mathbb{R} \to \mathbb{R}$ is injective if and only if it has no local extrema. (If you're stuck, it's discussed here.)

Assuming that statement is true, then the result is easy. Indeed, if $f: \mathbb{R} \to \mathbb{R}$ is continuous but not injective, then $f$ has a local extrema at some $x_0$, and it follows (show this) that $f$ is not locally injective at $x_0$.

Edit: I didn't notice the comments before posting this. The solution there is similar but more to the point.

Community
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Dan
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