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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`
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
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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
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Sep 30 at 16:51 history asked asv CC BY-SA 4.0