Timeline for The closure of the set of injective continuous functions
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2021 at 8:30 | vote | accept | ABIM | ||
| Sep 23, 2021 at 20:06 | answer | added | KP Hart | timeline score: 5 | |
| Sep 23, 2021 at 0:02 | history | edited | ABIM | CC BY-SA 4.0 |
edited body
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| Sep 22, 2021 at 17:41 | comment | added | Pietro Majer | In fact if n<m/2 I think by Sard's theorem one can make all self intersections transverse, which means no self-intersections | |
| Sep 22, 2021 at 17:36 | comment | added | Pietro Majer | If n<m/2 one can first approximate the map by a smooth one, and then try to approximate by injective ones. There should be enough room. | |
| Sep 22, 2021 at 17:35 | comment | added | Nate Eldredge | @TomTheQuant: Sure, in 3 or more dimensions you can make it injective, but the point is not in 2. | |
| Sep 22, 2021 at 17:31 | comment | added | Pietro Majer | I'd say there is the same obstruction if $n\ge m/2$: there is a smooth map with say $f(a)=f(b)=0$ and $df(a)\pitchfork df(b)$. Say the images of the differentials are two linear spaces whose sum is the whole space. Then I think any close enough continuous approximation of $f$ has a self-intersection too, by topological degree reasons. | |
| Sep 22, 2021 at 17:21 | comment | added | ABIM | @PietroMajer Does it though, or (if there are enough dimensions) can it look like $\otimes$ but it $\epsilon$-comes close to crossing itself at that point. | |
| Sep 22, 2021 at 17:15 | comment | added | Pietro Majer | just consider a nbd of the crossing point; say that there the curve to be approximated looks like this $\otimes$. Any approximating curve has to follow the two segments, and needs to cross itself. | |
| Sep 22, 2021 at 17:14 | comment | added | Jochen Wengenroth | @PietroMajer oops..., I don't follow either. | |
| Sep 22, 2021 at 17:11 | comment | added | ABIM | @PietroMajer Why is this? | |
| Sep 22, 2021 at 17:10 | history | edited | ABIM | CC BY-SA 4.0 |
added 700 characters in body
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| Sep 22, 2021 at 17:04 | comment | added | Pietro Majer | But note that a curve $\mathbb R\to\mathbb{R^2}$, whose image has a self-crossing point, can't be approximated by injective curves | |
| Sep 22, 2021 at 16:58 | comment | added | ABIM | @PietroMajer Ah true, for example $f_1\in C(\mathbb{R},\mathbb{R})$ is any strictly monotone increasing function. Then, for any $f_2\in C(\mathbb{R},\mathbb{R})$, the map $f:\mathbb{R}\rightarrow \mathbb{R}^2$ defined by $f(x)=(f_1(x),f_2(x))$ is injective (even if $f_2$ need not be). Takeaway: I it's sufficient for one component of a map $f\in C(\mathbb{R},\mathbb{R}^m)$ to be monotonically increasing but the injectivity of $f$ does not imply that all its components to be monotone). | |
| Sep 22, 2021 at 16:56 | comment | added | Pietro Majer | @JochenWengenroth I don't follow. If a curve is injective, its components need not to be injective. | |
| Sep 22, 2021 at 15:35 | comment | added | Jochen Wengenroth | I doubt that $\mathcal I(\mathbb R^n,\mathbb R^m)$ is ever dense in $C(\mathbb R^n,\mathbb R^m)$. For $n=1$ this should follow from the fact that every component of a continuous injective function $\mathbb R\to\mathbb R^m$ is strictly monotone. | |
| S Sep 22, 2021 at 15:16 | history | suggested | Dirk Werner | CC BY-SA 4.0 |
typo corrected
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| Sep 22, 2021 at 15:14 | review | Suggested edits | |||
| S Sep 22, 2021 at 15:16 | |||||
| S Sep 22, 2021 at 15:07 | review | First questions | |||
| Sep 22, 2021 at 15:14 | |||||
| S Sep 22, 2021 at 15:07 | history | asked | ABIM | CC BY-SA 4.0 |