Timeline for Is the complement of a square imbedded to a cylinder connected?
Current License: CC BY-SA 4.0
12 events
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Sep 30 at 21:09 | comment | added | Pietro Majer | C may be seen as a plane annulus. Joining the other 2 edges of the imbedded square and the arcs A\f(a) and A'\f(a') one gets a simple closed curve, and I'd say C\f(Q) is connected by the Jordan curve thm. | |
Sep 30 at 19:41 | history | edited | LSpice | CC BY-SA 4.0 |
`\backslash` -> `\setminus`
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Sep 30 at 19:08 | comment | added | Anton Petrunin | By Moore’s quotient theorem, if a continuous onto map $f$ from the sphere $\mathbb{S}^2$ to a Hausdorff space $X$ has acyclic fibers, then $f$ can be approximated by a homeomorphism; in particular, $X$ is homeomorphic to $\mathbb{S}^2$. It seems that your statement follows from this theorem, pass to the quotient with one-point set instead of each boundary component. | |
Sep 30 at 18:55 | comment | added | asv | @RyanBudney; Do you have a reference to this theory? A more focused information would be helpful. | |
Sep 30 at 18:53 | comment | added | Ryan Budney | Take a look at the theory of surfaces, and embedded curves in those surfaces. There is a basic theory of how cutting along embedded curves affects the surface: its genus and number of boundary components. You can derive the theory from the classification of surfaces, or simply from Poincare Duality. | |
Sep 30 at 17:36 | review | Close votes | |||
Sep 30 at 20:49 | |||||
Sep 30 at 17:25 | comment | added | asv | @YaakovBaruch: Sorry, corrected. | |
Sep 30 at 17:25 | history | edited | asv | CC BY-SA 4.0 |
edited body
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Sep 30 at 17:12 | comment | added | Aleksei Kulikov | I would assume it was supposed to mean $f(Q\backslash (a\cup a'))$... | |
Sep 30 at 17:11 | comment | added | Yaakov Baruch | What does $f(\partial C)$ mean? Isn't $f$ defined on $Q$? | |
Sep 30 at 17:00 | history | edited | asv | CC BY-SA 4.0 |
edited body
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Sep 30 at 16:51 | history | asked | asv | CC BY-SA 4.0 |