Can two convex polygons homeomorphic ? I know that the boundary of a polygon is homeomorphic to the circle $S^{1}$ but don't know with the interior ? Anybody help me ?
Two convex polygons can be homeomorphic?
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general-topology
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0What is the definition of a convex polygon (that you are working with)? As you are asking to make the homeomorphy specific (in the comments below ), it would help to have an explicit definition. – 2017-04-19
3 Answers
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Any convex polygon is homeomorfic to the unit ball (go to the polar coordinates with the centre in the inner point)
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1Can you write down it clearly ? – 2017-02-24
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0Let inner point is center of (r,f) - polar coordinates. G(f) is a maximal radius by angle f. See (r,f) -> (f/G(f), f) – 2017-02-24
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0I know the way to prove the boundary of a covex polygon is homeomorphic to $S^{1}$ . The first I can assume polygon lie on circle , so each point $a$ on boundary of polygon is correspond to a point which is the intersection between the line through center-the point $a$ and the circle ( call this homeomorphism is $h$ between two boundary of covex polygon ), then I can extend this homeomorphism by if $p,q$ are two points in polygonal regions , respectively . Then the line segment from $p$ to $x$ is correspond to the line segment from $q$ to $h(x)$ . Is this prove true ? – 2017-02-24
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0?? Sorry for ask again ? – 2017-02-24
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If you already know that boundary of convex polygon is homeomorphic to $S^1$ then you can easily see that any solid convex polygon (by solid I just mean that we look both at it interior and boundary) is homeomorphic to the disk $D^2$ (and $\partial D^2 = S^1$). Since any two solid convex polygons are homeomorphic to $D^2$ they are homeomorphic to each other too.
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0Yes , but I don't know how to prove with the interior ? – 2017-02-24
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1You can write the direct formula for homeomorphism as kotomord suggested but in such simple case maybe it will be even better to imagine this topologically. Two spaces are homeomoprhic if you can deform one to another without ripping into pieces (you are able to blow, squeeze, scale and deform as you want but can't cut the material). Suppose that you have the disk inscribed into the square, then you can blow it up to the whole square. It works similarly with polygons too. – 2017-02-24
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0I know the way to prove the boundary of a covex polygon is homeomorphic to $S^{1}$ . The first I can assume polygon lie on circle , so each point $a$ on boundary of polygon is correspond to a point which is the intersection between the line through center-the point $a$ and the circle ( call this homeomorphism is $h$ between two boundary of covex polygon ), then I can extend this homeomorphism by if $p,q$ are two points in polygonal regions , respectively . Then the line segment from $p$ to $x$ is correspond to the line segment from $q$ to $h(x)$ . Is this prove true ? – 2017-02-24
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0Why do you assume that boundary of any random polygon lies on a circle? It is true only for some kind of "good" polygons like triangle, square and so on. Or maybe I don't understand you? You may visually see that the boundary of polygon is homeomorphic to circle because circle can be obtained from polygon by smoothing all corners, or, saying it in another way, that $S^1 = \lim\limits_{n\to\infty} P_n$, where $P_n$ is a polygon with $n$ corners. – 2017-02-24
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0Because two $n$ polygon always homeomorphic , so I can assume it lie on circle – 2017-02-24
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0If you understand this fact then it's OK. Yes, you can just draw lines from circle center and send points of polygon's boundary to circle by $x_1 \to x_2$, where $x_1$ corresponds to intersection of line with the boundary and $x_2$ to line with the circle. – 2017-02-24
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Another suggestion that you can use the Riemann mapping theorem to see that any two convex open polygons (i.e. simply connected open subset of the complex plane) are conformally equivalent, which means the map between them is not only homeomorphic but also conformal. Ref: https://en.wikipedia.org/wiki/Riemann_mapping_theorem