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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?

Aryabhata
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Aleksei Averchenko
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    What about a Hilbert curve? – jimjim Jun 25 '11 at 06:58
  • Chris Eagle gave an answer here, using a completely different route. Your suspicion is right, by the way. – t.b. Jun 25 '11 at 06:58
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    @Arjang: Those are not injective. – t.b. Jun 25 '11 at 06:59
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    @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? – jimjim Jun 25 '11 at 07:08
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    @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 07:14
  • @Theo : I was looking for a key concept to ponder rather than wether just a special case, I rather look at why unit interval and unit square are not homeomorphic, thanks to your suggestion. – jimjim Jun 25 '11 at 07:22
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    http://arxiv.org/pdf/1003.1467.pdf – user27182 Mar 26 '13 at 01:00

3 Answers3

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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.

Zarrax
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  • Nice answer! But let me ask you one question: so you said that $f^{-1}(B)=I$ where $I$ is closed interval. It is closed because $f^{-1}$ is continuous and $B$ compact, right? Is it possible that $I=[a,a]$? – RFZ Nov 08 '19 at 04:56
  • Btw, where you are using that $f$ is injective and surjective? – RFZ Nov 08 '19 at 05:02
  • $I$ can't be $[a,a]$ since $f^{-1}$ is one to one; $B - {x}$ has more than one point, so $f^{-1}(B - {x})$ also has more than one point. As for where we are using injectivity and surjectivity ... even to talk about $f^{-1}$ requires one to oneness, for example. Surjectivity is used when saying $\cup_n A_n = {\mathbb R}^n$. – Zarrax Nov 09 '19 at 23:45
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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!

Juan S
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  • It's a cylinder, of course. – Aleksei Averchenko Jun 25 '11 at 07:29
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    @Alexei - keep homotoping! – Juan S Jun 25 '11 at 07:30
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    A circle then? :) – Aleksei Averchenko Jun 25 '11 at 07:31
  • @Alexei - indeed. – Juan S Jun 25 '11 at 07:32
  • 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? – Aleksei Averchenko Jun 26 '11 at 01:10
  • @Alexei - homeomorphic spaces have isomorphic homology groups. Removing a point preserves the homeomorphism. But the homology groups are different, so they could not have been homeomorphic originally – Juan S Jun 27 '11 at 08:48
  • But homeomorphism isn't claimed, only continuous bijection is. It could still sort of work if (co)homology turned mono into mono, but I'm unaware if that's the case. – Aleksei Averchenko Jun 27 '11 at 09:08
  • @Alexei - of course you are correct, sorry I was just reading it as homeomorphic. I am sure there is a way to salvage this proof, but I can't see it immediately. I'll have a think, hopefully someone will post in the meantime – Juan S Jun 27 '11 at 23:44
  • What does $0$ mean in the first diagram? $\mathbf{Top}$ doesn't have a zero object, and if you meant the initial object ($\varnothing$), then the last arrow just doesn't exist! Also, how do you define exactness without zero morphisms? Are you considering some subcategory of $\mathbf{Top}$ that has these things? – Aleksei Averchenko Jun 29 '11 at 07:26
  • @Alexei - opps, I will edit. The last few months I have strictly worked in abelian categories, and adding in zeroes as I need :) – Juan S Jun 29 '11 at 07:34
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From the Wikipedia Space-filling curves page:

A non-self-intersecting continuous curve cannot fill the unit square because that will make the curve a homeomorphism from the unit interval onto the unit square (any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism). But a unit square has no cut-point, and so cannot be homeomorphic to the unit interval, in which all points except the endpoints are cut-points.

Ste_95
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