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I need a hint. The problem is: is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$

I'm pretty sure that there aren't any, but so far I couldn't find the proof.

My best idea so far is to consider $f' = f|_{\mathbb{R}-\{*\}}: \mathbb{R} - \{*\} \to \mathbb{R}^2 - \{f(*)\}$, and then examine the de Rham cohomologies: $$H^1_{dR}(\mathbb{R}^2 - \{f(*)\}) = \mathbb{R} \ \xrightarrow{H^1_{dR}(f')} \ 0 = H^1_{dR}(\mathbb{R} - \{*\}),$$ but so far I failed to derive a contradiction here. Am I on the right path? Is it possible to complete the proof in this way e.g. by proving that $H^1_{dR}(f')$ must be a mono? Or is there another approach that I missed?

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What about a Hilbert curve? – Arjang Jun 25 '11 at 6:58
@Arjang: Those are not injective. – t.b. Jun 25 '11 at 6:59
@Theo : Thanks, I always imagined the limiting case of hilbert curve to be a mapping from $\mathbb R to \mathbb R^2$ , I feel disillusioned that is not the case. is it a question worth asking? – Arjang Jun 25 '11 at 7:08
@Arjang: It's your call. But why don't you try looking into it yourself? Usually they are from the unit interval onto to the unit square and if such a map were injective, it would be a homeomorphism, which it can't be for reasons of connection (removing an interior point of the interval disconnects it, whereas ...) – t.b. Jun 25 '11 at 7:14
2 – user27182 Mar 26 '13 at 1:00

2 Answers 2

up vote 19 down vote accepted

Suppose $f(x)$ were such a function. Note that each $A_n = f([-n,n])$ is a closed (actually compact) set, with $\cup A_n = {\mathbb R}^n$. By the Baire category theorem, there is one such $A_n$ that contains a closed ball $B$. Since $[-n,n]$ is compact, the image of any relatively closed subset of $[-n,n]$ is compact and thus closed. Hence $f^{-1}$ is continuous when restricted to $A_n$, and thus when restricted to $B$. So in particular $f^{-1}(B)$ is a connected subset of ${\mathbb R}$. Since all connected subsets of ${\mathbb R}$ are intervals, $f^{-1}(B)$ is a closed interval $I$.

Let $x$ be any point in the interior of $B$ such that $f^{-1}(x)$ is not an endpoint of $I$. Then $B - \{x\}$ is still connected, but $f^{-1}(B - \{x\})$ is the union of two disjoint intervals, which is not connected. Since $f^{-1}$ when restricted to $B - \{x\}$ is continuous, you have a contradiction.

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Here is a hint: What simple space is $\mathbb{R}^2 - \{ f(\ast) \}$ homotopic to?

Edit: Just a small edit, to hopefully bump this up. I had read this as a homeomorphism, in which case it is easy. However we only have a continuous bijection from $\mathbb{R} \to \mathbb{R}^2$.

There may be a way to argue from the fact that $\mathbb{R} - \{ \ast \}$ is disconnected and $\mathbb{R^2} - \{ f(\ast) \}$ is connected. This will work immediately to show there is no continuous bijection from $\mathbb{R}^2 \to \mathbb{R}$ as the continuous image of a connected set is connected. I am not sure about getting something out of the other direction however (perhaps the 'simplest' is Zarrax's explanation). Hopefully the experts will have something to add!

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It's a cylinder, of course. – Alexei Averchenko Jun 25 '11 at 7:29
@Alexei - keep homotoping! – Juan S Jun 25 '11 at 7:30
A circle then? :) – Alexei Averchenko Jun 25 '11 at 7:31
@Alexei - indeed. – Juan S Jun 25 '11 at 7:32
Can I have another hint? Am I to use $f'$ to construct a homotopy from $\mathbb{R} - \{*\}$ to $S^1$ or some kind of 'homotopic injection' that would translate to a mono from $H_0(\mathbb{R} - \{*\}) = \mathbb{Z}^2$ to $H_0(S^1) = \mathbb{Z}$ and thus get a contradiction, or am I to play with $f'$ as an attaching map, or is it something I still fail to see? – Alexei Averchenko Jun 26 '11 at 1:10

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