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Do there exist continuous bijections from Euclidean space $X: \mathbf R^n\to M$ whose inverse is not continuous (where $M$ is a $n$ dimensional manifold)?

I'm aware of continuous bijections from subsets of Euclidean space which change the topology, for example $\exp(2\pi i x):[0,1)\to \mathbf S^1$. However without a boundary I can't imagine how space can be wrapped up to change the topology. Hence the question.

The way one usually sets up a discontinuous inverse is to map faraway points nearby. To avoid messing up the manifold structure, these faraway points must be from infinity or the boundary. Euclidean space has no boundary so only the points at infinity can be used. But, there must be also be a point which bridges the points at infinity. This point can't exist because Euclidean space has no boundary. Hence the question.

dennis
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  • I don‘t understand your question, you‘ve given the standard example that answers your question yourself already. – Qi Zhu Sep 29 '22 at 14:29
  • @QiZhu Maps from the entirety of Euclidean space. – dennis Sep 29 '22 at 14:32
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    A better example to consider than your initial example is a map $f\colon (0,1)\to S^1$ where one end curves back and touches itself in the middle. – Cheerful Parsnip Sep 29 '22 at 14:38
  • @CheerfulParsnip Does such a map exist? – dennis Sep 29 '22 at 14:39
  • Definitely. Just imagine capital letter P. The curve starts at the bottom, continued up to the top, turns right, and comes back to join itself in the middle. To get an explicit example, you could take any curve that loops back on itself and clip the domain just right. – Cheerful Parsnip Sep 29 '22 at 14:44
  • @CheerfulParsnip Yes but P is not $\mathbf S^1$ – dennis Sep 29 '22 at 14:46
  • @CheerfulParsnip P is not a manifold. I guess it is not possible if we require $M$ to be a manifold. I wonder if there is a theorem for this. – dennis Sep 29 '22 at 14:51
  • @dennis Oh I see. In my head, I was thinking you meant a bijection onto its image so that $M$ could be $\mathbb R^2$ in this case. – Cheerful Parsnip Sep 29 '22 at 14:55

1 Answers1

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Since you require the image to be an $n$-dimensional manifold, there is no such map. This follows immediately from the invariance of domain theorem (proved using techniques of algebraic topology), which states that any continuous injective map between manifolds without boundary of the same dimension is an open map (and therefore a homeomorphism onto its image).

As pointed out in the comments, if you don't require the image to be a manifold, then there are lots of examples, for example by just looping $(0, 1)$ back on itself so the image is topologically a 6 shape (the symbol 6, not the number it denotes).

Daniel Hast
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  • Thanks. Is it possible for $M$ to have boundary? – dennis Sep 29 '22 at 15:09
  • Also do you have a reference for this version of the IDT theorem. Most of them use compactness. – dennis Sep 29 '22 at 15:20
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    I think a boundary shouldn't be a problem, just "extend" past the boundary a bit using the collar neighborhood theorem to get an $n$-dimensional manifold-without-boundary $M'$ in which $M$ is a closed subset (think replacing $(0, 1]$ with $(0, 1 + \varepsilon)$), then apply invariance of domain to this to see the image is open in $M'$ and hence can't be all of $M$. – Daniel Hast Sep 29 '22 at 15:28
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    Doesn't the version for topological manifolds just follow from working locally and applying the classical version for open subsets of Euclidean space? I don't see where compactness would be needed. – Daniel Hast Sep 29 '22 at 15:49